Dls 212 - Selected Topics in General Physics: University of Northern Philippines

Republic of the Philippines
UNIVERSITY OF NORTHERN PHILIPPINES

Vigan City, Ilocos Sur
DLS 212 – SELECTED TOPICS

IN GENERAL PHYSICS
Submitted by:
CHERRYL R.TOBIAS
OPEN UNIVERSITY
DLS 212 – Selected Topics in General Physics
MODULE 1
LESSON 1. LINEAR MOTION
LESSON 2. NONLINEAR MOTION

OPEN UNIVERSITY
CHERRYL R. TOBIAS
MAEd-Science Education
Lesson 1:
Self- Test 1
Answer the following questions:

1. Give a very brief and simple definition of the following terms:
ANSWER:
A. Displacement- it refers to the shortest distance between the object’s two position, like the
distance between its point of origin and its point of destination.
B. Velocity- speed with direction.
C. Acceleration- change of velocity over a period of time.
2. A car travels 60 km/hr to the south in a straight line. What is the speed of the car? Its velocity?
How far did it go after 1 hour?
ANSWER:
Given: Speed= 60km/hr Velocity= 60km/hr south
a) Speed of the car?
* As the situation given speed of the car is 60km/hr therefore S= 60km/hr
b) Its velocity?
* As the situation given velocity of the car is 60km/hr to the south therefore V= 60km/hr
South
c) How far did it go after 1 hour?
* Speed= Distance/Time
Speed = (60 km/hr)
Time= (1hr)
Distance= ?
60 km/hr = Distance/ 1 hour
Distance = 60km travelled after 1hour
3. Alfred John recorded the time it takes a calesa to cover 10 meters. The data are as follows:
Distance from start Time(s)

(m)
10 2
20 40
30 61
40 82
50 104
60 124
70 146
80 169
90 193
100 216
a. Describe the motion of the calesa (How did the calesa change with time? Show graphically
how the position changes with time.)
Answer:
Formula Solution
 20m t= t2- t1 40-2=38 sec.

 30m t= t3-t2 61-40= 21 sec.
 40m t= t4-t3 82-62= 20 sec.
 50m t= t5-t4 104-82= 22 sec.
 60m t= t6-t5 124-104= 20 sec.
 70m t= t7-t6 146-124= 22 sec.
 80m t= t8-t7 169-146= 23 sec.
 90m t= t9-t8 193-169= 24 sec.
 100m t= t10-t9 216-193= 23 sec.
Show graphically how the position changes with time.)
100
90
80
D 70
i
60
st
a 50
n
c 40
e 30
20
M
10
0
20 40 60 80 100 120 140 160 180 200 220 240
Time (seconds)
B. Was the calesa accelerating? (Show graphically hot its velocity changes with time.)
Answer:
V= X/ t = x2-x1/ t2-t1
 10m v= 10m/ 2 sec = 5m/sec

 20m v= 10m/ 38 sec = 0.26 m/sec
 30m v= 10m/ 21 sec = 0.48 m/sec
 40m v= 10m/ 21 sec = 0.48 m/sec
 50m v= 10m/ 22 sec = 0.45m/sec
 60m v= 10m/ 20 sec = 0.50m/sec
 70m v= 10m/ 22 sec = 0.45m/sec
 80m v= 10m/ 23 sec = 0.43m/sec
 90m v= 10m/ 24 sec = 0.42m/sec
 100m v= 10m/ 23 sec = 0.43m/sec
(Show graphically hot its velocity changes with time.)
5
V 4
e
l 3
o
2
c
i 1
t
y 0
20 40 60 80 100 120 140 160 180 200 220 240
Time (seconds)
4. From the graphs given below, describe the acceleration of a particle.

a) b)
x v
t t
Answer:
a. The acceleration is zero as you can see in the graph the distance is proportion to the time, so
it means the velocity of the object is contrast.
b. The acceleration is positive as you can see in the graph the velocity is increasing as time
passes.
5. An airplane makes a straight back and forth round trip, always at the same airspeed, between two
cities. If it encounters a mild steady tailwind going, and the same steady headwind returning, will the
roundtrip take more, less or the same time as with no wind?
Answer:
A and B = Two cities
= Direction of speed SPEED= Distance/ Time
*Distance is constant so Speed= Time
TIME= SPEED (Situation 1 with no wind)
GOING BACK
A S ( Airplane) B A S ( Airplane) B
Time (Total) = Time travel from A To B + Time travel from B to A

Time= s Airplane + s Airplane
= 2 s Airplane
SITUATION 2 with Wind
GOING BACK
A S ( Airplane) B A S ( Airplane) B
S ( wind) S ( wind)
Time (Total) = Time travel from A to B + Time travel from B to A

Time= S (airplane) + S (airplane)
Time = (S airplane + S wind) + (S airplane – S wind)
= 2 S airplane
Therefore: The travel will take same time, as with wind or no wind.
6. What is the average speed of the Calesa (Refer to prob.3)

Answer:
Average Speed= Distance Total Covered/ Time
= 100 m/ 216 S
= 0.46 m/s
7. On May 28, 2000, Juan Montoya became the first Columbian citizen to wind the Indianapolis 500.
Montoya completed the race in a time of 2.98 j. what was her average speed during the 500 mi race?
Answer:
Find: Average Speed during the 500 mi race?
Given: Distance Covered= 500 mi Time 2.98 j
Average Speed= Distance Covered/ Time
= 500 mi/ 2.98 j
= 167. 79 mi/j
8. The world speed record on water was set on October 8, 1978 by Ken Warby of Blowering Dam,
Australia. If Ken drove his motorboat a distance of 1000 m in 7.045 s, how fast was his boat moving
a) in m/s? b) In mi/h?
Answer:
Given: Distance= 1000 m Time= 7.045 sec
Find: a. SPEED in m/s b. in mi/hr
Solution:
a. SPEED= Distance/ Time b. m/mi?
= 1000 m/ 7.045 sec = 141.94 m/sec x 1mi/1609.34 m x 3600 sec/ 1hr
= 141.94 m/sec =317.51 mi/hr
9. Hans stands at the rim of the Grand Canyon and yodels down to the bottom. He hears his yodel
echo back from the canyon floor 5. 20 s later. Assume that the speed of sound in air is 340. 0 m/s.
How deep is the canyon at this location?
Answer:
Given: Speed= 340 m/s Time= 5.20 S
Find: Distance=?
Distance= (340) (5.20) / 2
= 884 m
10. The slowest animal ever discovered was a crab found in the Red Sea. It traveled with an average
speed of 5.70 km/y. How long would it take this crab to travel 100. 0 km?
Answer:
Average Speed= 5.70 Km/y Distance Travel= 100 km
Find: How long would it take?
Solution: Average Speed= Distance Travel/ Time
5.70 km/y = 100km/ time
Time= 17.54 y
11. Grace is driving her sports car at 30 m/s when a ball rolls out into the street in front of her.
Grace slams on the brakes and comes to a stop in 3.0 s. What was the acceleration of Grace’s car?
Answer:
Given: Vi= 30m/s t= 3.0 s Vf= 0
Find: a=?
Solution: acceleration = Vf- Vi/ t
= 0. 30 m/s /3.0 sec
= -10 m/sec2
12. Benjamin watches a thunderstorm from his apartment window. He sees the flash of lightning bolt
and begins counting the second until he hears the clap of thunder 10 s later. Assume that the speed
of sound in air is 340 m/s. How far away was the lightning bolt a) in m? b) in km?
Answer:
Given: S= D/T S= 340 m/ t= 10 s
A. In m? B. in km
D= s (t) * conversion m to km
= (340 m/s) (10s) = 1000 m = 1km
= 3400 m D= 3400 m x 1 km/ 1000m
= 3.4 km
13. It is now 10:29 a.m. but when the bell rings at 10:30 am, Suzette will be late for French class for
the third time this week. She must get from one side of the school to the other by hurrying down
three different hallways. She runs down the first hallway, a distance of 35.0 m, at a speed of 3. 50
m/s. The second hallway is filled with students, and she covers its 48.0 m length at an average speed
of 1. 20 m/. The final hallway is empty, and Suzette sprints its 60.0 m length a speed of 5. 00 m/s.
A. Does Suzette make it to class on time or does she get detention for being late again?
B. Draw a distance vs. time graph of the situation.
Answer:
SPEED (m/s) DISTANCE (m)
1st Hallway s= 3. 5 m/s 35m
2nd Hallway v= 1.2 m/s 48 m
Find Hallway s= 5 m/s 60m
1st Hallway Speed = distance/time
3.5m/s 35 m/t
t= 10 sec
2nd Hallway Vav= S/T
1.2 M= 48M /t
T= 40sec.
Final Hallway speed= ds/t
5 m/s= 60-48 / t
T= 24 sec
14-15 incomplete
OPEN UNIVERSITY
CHERRYL R. TOBIAS
Lesson 1:
SELF TEST II
Answer the following problems:
1. Suppose you take a trip that cover 240 km and takes 4 hours to make. What is your average
speed?
Answer: Ave. SPEED = 240 km/ 4 hrs
= 60 km/hr.
2. If a car is moving in a straight line and steadily increase its speed. It moves from 35 km/ hr. to
40 km/hr. the first second and from 40 km/ hr. to 45 km/hr. the next second. What is the car’s
acceleration?
Answer:
GIVEN: Velocity: 35 km/hr. 40km/hr. 45km/hr.
Find: Accelerating of the car
Solution: Vx= Vxo + at
40= 35 + at
Equation #2 X= 35 t + ½ at
Equation #1 T= 5/a
X= Xo + V xo t + ½ at
X= 0 +35t+ ½ at
Equation #2 x=35t ½ at
Vx2= Vxo 2 + 2a (X- Xo)
40(2) = 35(2) + 2a (X)
1600= 1225 + 2a (35t + ½ at)
1600= 1225 + 70 at + a (2) t
1600= 1225 + 70 a ( 5/A) + a2 (5/a)
1600=1225 + 350 + 5a
25= 5a
A= 5km/Hr 2
3. A car accelerate at 2 m/s2. Assuming the car starts from rest, how much time does it need to
accelerate to a speed of 30 m/s.?
Answer: Given: a= 2 m/s2 Vx= 30m/s t=?
Solution= Vx= Vxo + at
30= 0 + 2t
T= 15 sec
4. Ten seconds after staring from rest, what will be the speed of a freely falling body?
Answer: Given: t=10 sec g= 9.8 m/s 2
Find: Vy=?
Vy= Vyo – gt
Vy= 0- (9.8) (10) = -98 m/s
5. If an object were equipped with a speedometer and allowed to fall freely on a planet where
the acceleration due to gravity is 20 m/s2, what would be the reading of the speedometer?
Answer: Given: g= -20 m/s 2
Find: Reading of the speedometer
Solution: Vy= Vyo- gt
Vy= 20 m/s2 (t)
6. Vivian is walking to the hairdressers at .3 m/s when she glances at her watch and realizes that
she is going to be late for her appointment. Vivian gradually quickens her pace at a rate of
0.090 m/s2
A. What is Vivian’s speed after 10.0 S?
B. At this speed, is Vivian walking, jogging, or running very fast?
Answer: Given: Vxo= 1.3 m/s a= 0.090 m/s 2
a. Vx= Vxo + at b. SHE IS RUNNING VERY FAST
Vx= 1.3 + (0.090) (10)
Vx= 2.2 m/s
7. King Kong carries Far Wray up the 321 m- tall Empire state building. At the top of the
skyscraper, fray ways show falls from her foot. How fast will the shoe be moving when it hits
the ground?
Answer: Given: Y= 32/m g= 9.8 m/s2 Vy=0
Find: Vy=?
Solution: Vy2= Vy0 (2) – 2g (y-yo)
Vy2= 0(2) - 2(9.81) (321-0)
Vy= -79.32 m/s
8. Chief Boolie, the jungle dweller, is out hunting for dinner when a coconut falls from tree and
lands on his toe. If the nut fell for 1.4 s, how fast was it travelling when it hit chief Boolie’s
toe?
Answer: G= 9.8 m/s 2 t= 1.4 s
Vy= Vgo – gt
Vy= - (9.8) (1. 4)
= -13.72 m/s
9. How long Tina, a ballerina, in the air when she leaps straight up with a speed of 1.8 m/s?
Answer:
Given: Vyo= 1.8 m/s g= 9.8 m/s2
Vy- Vyo – gt Tina time in the Air
0= 1.8- (9.8) t = 0. 184 (2)
9.8 t = 1.8 = 0.368 seconds.
t= 0.184 sec
10.In order to open he clam it catches, a seagull will drop the clam repeatedly onto a hard
surface from high in the air until the shell cracks. If a seagull flies to the height of 25 m, how
long will the clam take to fall?
Answer:
Given: y= 25 m g= 9.8 m/s2
Y= Yo + Yot- ½ gt 2
25= -1/2 (9.8) m/s2
T= 2. 26 sec
11.The steamboat geyser in Yellowstone National Park, Wyoming is capable of shooting its hot
water up from the ground with a speed of 48.0 m/s. How high can this geyser shoot?
Answer:
Given: Vy=48 m/s g= 9.8 m/s2
O= 48 – (9.8) (T) y= Yo + Y o t – 1/2g t 2
T= 4.9 sec y= 48 (4.9) - ½ (9.8) (4.9) 2
= 117. 55 m
12.At six flags great adventure Amusement park in New Jersey, a popular rye known as “free fall”
carries passengers up to a height of 33.5 m and drops them to do the ground inside a small
cage. How fast are the passengers going at the bottom of this exhilarating journey?
Answer:
Given: Vy2= Vyo2- 2g(Y-Yo)
Vy2= -2 (9.8) (33.5)
Vy= -25.62 m/s
14. In the wizard of Oz, Dorothy awakens in Munchkinland where her house has been blown by a
tornado. If the house fell from a height of 3000m, with what speed did it hit the wicked witch
of the East when it landed.
Answer:
Given: y=3000 m g= 9.8 m/s2
Vy2= Vyo 2- 2g (y-Yo)
Vy2= -2 (9.8) (3000)
Vy= - 242. 49 m/s
15.A baby blue jay sits in a tall tree awaiting the arrival of its dinner. As the mother lands on the
nest, she drops a worm toward the hungry chick’s mouth, but the worm misses and fall from
the nest to the ground at 1. 50 s, how high up is the nest?
Answer:
Given: Y= Yo + V yo t- 1/2gt 2
Y= -1/2 (9.8) (1.5) 2
= 11. 025 m
OPEN UNIVERSITY
CHERRYL R. TOBIAS
Lesson 1:
SELF TEST III
1. At the instant a horizontally held rifle is fired over a level range, a bullet held on the side of
the rifle is released and drops to the ground. Which bullet, the one fired downrange or one
dropped from rest, strikes the ground first?
Answer: The two bullets strike the ground at the same time.
2. At what part of its trajectory does a projectile have

a. Minimum speed
b. Maximum speed?
Answer:
a. minimum speed- when the projectile is at lowest
b. maximum speed- when the projectile is at highest
3. Discuss the difference between linear speed and rotational speed.

Answer:
Rotational speed is the rate of change of the angle (in radians) with time, while linear speed is the
speed of a point on the surface of the spinning object, which is the angular speed times the distance
from the point to the axis of rotation.
4. Differentiate rotation from revolution.
Answer:
Rotation is the motion of a body in which the body turns around its own axis without any change
in its position while revolution is the movement of a body along a circular path with continuous
change in its position.
5. If St. Louis Cardinals homeroom king, Mark Mcgwire, hit a baseball due west with a speed of
50.0 m/s, and the ball encountered a wind that blew it north at 5. 00 m/s, what was the
resultant velocity of the baseball?
Answer: N
5m/s V=? V= 50 + 5(2)
W E = 50.25 m/s
50m/s
S
6. Ivan pulls a sled loaded with logs to his cabin in the woods. If Ivan pulls with a force of 800 N
in a direction 20.0 above the horizontal, what are the horizontal and vertical components of
the force exerted by Ivan?
Answer: Y
800N = f FIND: Fx and Fy
20 X Fx= F cos 0 Fy= F sin 0
Fx= 800 cos 20 = 800 sin 20
= 751.75 N = 273. 62 N
7. In her physics lab, Melanie rolls a 10g marble down a ramp and off the table with a horizontal
velocity of 1.2 m/s. the marble falls in a cup placed 0.51 m from the table’s edge. How high is
the table?
Answer: x= Vx (t)
0.51 = 1.2 t
T= 0. 425 sec
y= Yo Yo t – ½ gt 2
y= -1/2 (9.8) (0. 425) 2
= 0. 885 m
8. Jack be nimble, jack be quick, jack jumped over the candlestick with a velocity of 5.0 m/s at
an angle of 30.0 to the horizontal. Did jack burn his feet on the 0.25 m- high candle?
Answer: Vy= Vyo- gt Y= Yo + V yo t – ½ g t 2
0= 5 sin 30 – 9.8 t y= 5 sin 30 (0.2555) – ½ (9.80) (0.255) 2
T= 0.255 sec Y= 0.32 m 0.25m - Height of candle
Therefore jack will not burn his feet
9. Bert is standing on a ladder picking apples in his grandfather’s orchids. As he pulls each apple
off the tree, he tosses it into a basket the tree, he tosses it into a basket that sits on the
ground 3.0m below at horizontal distance of 2.0m from Bert. How fast must Bert throw the
apples (horizontally) in order to land in the basket?
Answer:
X= V x t
Y= Yo + V y o t – ½ g t 2
3m = -1/2 (9.8) (t) 2
T= 0.782 sec
X= V x t
2m = V x (0.782)
Vx = 2. 55 m/s
10.Emmanuel Zacchini, the famous human cannonball of the Ringing Bros, and Barmum and
Billey Circus, was fired out of a cannon with of 24.0 m/s at an angles 40.0 to the horizontal. If
landed in a net 56.6 m away at the same height from which he was fired, how long was
Zacchini in the air?
Answer:
t A-C = 24 sin 0 / g
t A-C = 2(24) sin 40 degree / 9.8 m/s
= 3.15 sec.
11.Caitlin is playing tennis against a wall. She hits the tennis ball from a height of 0.5m above the
ground which a velocity of 20.0m/s at the angle of 15.0 to the horizontal toward the wall that
is 6.00m away.
a. How far off the ground is the ball when it hits the wall?
b. Is the ball travelling up or is it on its way down when it hit the wall?
Answer:
X= V x t x= V x t
Vx= Vcos 0 6m= 19.32 (t)
Vx= 20 cos 15 degree t= 0.31 sec
Vx= 19.32 m/s
Vy= Vy o – gt
Vy= V sin 0 - gt
Vy= 20 sin 15 degree (9.8)(0.31 sec)
Vy= 2.138 m/s
Vy2= Vyo (2) – 2 g (y-y)
2.138 (2) = 5. 176 (2) – 2 (9.8) (y)
Y= 1.13m
A. Height from the ground B. the value of the Vy when hit the wall is positive
= 0.5 +y
=0.5 + 1.13 therefore the ball is travelling up.
= 1. 63m
12. At the fermi lab particle accelerator in Batavia, illios, protons are accelerated by
electromagnets around a circular chamber of 1.00 km radius to speeds near the speed of light
before colliding with a target to produce enormous amounts of energy. If a proton is travelling
at 10% the speed of light, how much centripetal force is exerted by the electromagnets?
(Hint: the speed of light is 3.00 x 10 (8) m.s, mp= 1.67 x 10(-27kg)
Answer:
Given: V= 10% (3x 10to the 8th power m/s) = 3x 10 to the 7th power m/s
R= 1 km m= 1. 67 x 10 (-27th power) kg. FC=?
FC= M V2/ r
= 1.67 x 10(-27th power) (3x10(7th power)2 / 1000
= 1. 503 x 10 (-15th power) kg.m/s2
13.To test their stamina, astronauts are subjected to many rigorous physical test before they fly
in space. One such test involves spinning the astronauts in a device call centrifuge that
subjects them to accelerations far greater than gravity. With what linear speed would an
astronaut have to spin in order to experience an acceleration of 3g’s at a radius of 10.0 m?
(1g=10.0 m/s 2)
Answer:
Given: g= 10 m/s2 r= 10 m Find: V=?

Ac= 3g-3 (10m/s2)= 30 m/s2
Ac= V2/r
30 m/s2 = V2/10m
V= 17.32 m/s
14.Roxanne is making a strawberry milkshake in her blender. A tiny, 0.0050 kg strawberry is

rapidly spun around the inside of the container with a speed of 14.0 m/s, held by a centripetal
force of 10.0 N. What is the radius of the blender at this location?
Answer:
Given: Fc= 10 N M= 0.0050kg V= 14.0 m/s
Find: r=?
Fc= m v2/r
10N = 0.0050 kg (14m/s)2 / r
R= 0.098 m
15.A popular trick of many physics teachers is to swing a pail of water around in a vertical circle
fast enough so that the water doesn’t spill out when the pill is upside down. If Mr. Lowell’s
arm in 0.60 m long, what is the minimum speed with which he can swing the pail so that the
water doesn’t spill out at the top of the path?
Answer:
The rotational speed create by the arm is equal and not greater to the tangential speed.
OPEN UNIVERSITY
CHERRYL R. TOBIAS
Lesson 1:
ASSIGNMENT:
1. What is time rate of change? Give three examples.
Answer:
Time rate of change is a quantity divided by time. Common examples are speed, velocity and
acceleration.
2. Distinguish between speed and velocity.
Answer:
Speed is the rate at which an object is moving at a certain distance over a period of time while
velocity is rate at which an object is moving a certain distance over a period of time with the
consideration of its direction.
3. If you were standing in an enclosed car moving at constant velocity, would you learned in
some special way to compensate for the car’s motion? What if the car were moving with
constant acceleration? Explain.
Answer:
No, because the velocity does not change so is the speed. My travel is so smooth.
4. On july 31, 1994, Sergey Bubka of Ukraine broke his own pole- vaulting record by attaining a
height of 6.14m.
a. How long didi it take Bubka to return to the ground from the highest part of his vault?
b. Describe how this time compares to the time it took him to go from the ground to the highest
point.
ANSWER:
a. t=? b. The time is the same as he going up and going down back
Y= Yo + V y o t – ½ g t 2 to the ground. Because as he start, the velocity at ground is decelerating
Y= ½ g t 2 until the velocity become zero and start getting down. And also the velocity
6.14= ½ (9.8)(t2) is changing and giving up because of its acceleration due to gravity until he
T= 1.12 sec reached the ground.
5. A projectile is launched upward at an angle of 750 from the horizontal and strikes the ground a
certain distance downrange. What the other angle of launch would produce the same distance?
ANSWER:
R1 = V2 sin /g
R2= v2 sin 2 (90 - )/g
R2= v2 sin (180-2 ) /g
But sin 180 = 0 therefore
R2 = V2 sin 2 /g
Therefore: 75
90 - = 90- 75 = 15
6. A seal swims toward an inlet with a speed of 5. 0 m/s as a current of 1.0 m/s flows in the opposite
direction. How long will it take the seal to swim 100 m?
ANSWER: 5.0 m/s 1m/s
V= d/t d= 100 m/t
5.1 = 100 m/ t
T= 25 secs
7. If the rotational speed of a platform is doubled, how does the linear, speed anywgere on the
platform change?
ANSWER:
If the rotational speed is double, the tangential speed is also doubled. Rotational speed is
related to tangential speed.
8. Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly for lunch. If the lily
pads are spaced 2.4 m apart, and Ferdinand jumps with a speed of5.0 m/s, taking 0.60 s to go from
lily pad to lily pad, at what angle must Ferdinand make each of his jumps?
ANSWER: R= V cos t
2. 4 = 5 cos (0.60)
= 36.87
10. Jessica is riding on a merry – go – round on an outer horse that sits at a distance of 8.0 m from
the center of the ride. Jessica’ sister, Julie, is on an inner horse located 6.0 m from the rider’s center,
the merry-go-round turns around once every 40.0 s.
a. Explain which girl is moving with the greater linear speed.
b. What is the centripetal acceleration of Julie and her horse?
ANSWER:
Location: From Center Jessica r= 8m Julie r= 6 m
W= 1 rotation/ 40 sec A. Jessica has the greater linear speed

W= V/r B. Ac= v2/r
1/40 = v/8m Ac= 0.15 (2) / 6m
V= 0.2 m/s (Jessich) Ac= 3.75 x 10 (to the -2nd power) m/s 2
1/40 = v/ 6m
V= 0.15 m/s julie

OPEN UNIVERSITY
MODULE 2
LESSON 1. FORCE
LESSON 2. FIRST LAW OF MOTION
LESSON 3. SECOND LAW OF MOTION
LESSON 4. THIRD LAW OF MOTION

OPEN UNIVERSITY
CHERRYL R. TOBIAS
Lesson 1:
MODULE 2
SELF- TEST 1
A. MODIFIED TRUE OR FALSE.

Write T on the line for the sentences that are True. If you decide that the sentence is false, change
the word/s in italics to make the sentence true.
Force 1. Gravitation is any push or pull that affects matter.
False (force of gravity) 2. The pull of Earth on a body is the body’s weight.
True 3. Newton is the unit used to measure force in the metric system.
False (Opposite direction) 4. Friction I a force that slows down an object by acting in a
direction along the motion of the object.
False (net force or resultant force) 5. The sum of the forces acting on an object is the total
force.
B. Give what are asked in the ff.

1. List down the following and distinguish one from the others.
a. The four fundamental forces in nature
b. Particular forces
Answer:
A) Gravitational force- covers the force due to gravity and weight.
Electromagnetic force- exerted by one atom on another atom.
Weak force occurs in radioactive decays
Strong force is the force that binds the constituents of protons and neutrons together.
B. Frictional force- is retarding force and always acts to slow down a moving object and it is
always directed opposite the motion of the object
Weight is force caused by the pull of a planet
Normal force- is a force exerted by a surface on an object.
Tension is the force exerted by the strings/rope
2. Explain the difference between mass and weight.
Answer:
Mass is the total amount of matter present in abject while weight is the force exerted by two
interacting object with masses. Mass is constant everywhere while weight is dependent on the pull of
gravity.
3. In general, the normal force is not equal to the weight. Give an example where the two forces
are equal in magnitude, and at least two examples when they are not.
Answer:
4. A car is driven up a steep hill at a constant speed. Discuss all the forces acting on the car.
What pushes it up the hill?
Answer:
5. To push a box up a ramp, is the force required smaller if you push horizontally of if you push
parallel to the ramp. Why?
Answer:
6. Maricel shovels snow after a storm by exerting a force of 30.0 N on her shovel at an angle of
60.0 degree to the vertical. What are with a force of the horizontal and vertical components of
the force exerted by Marcie?
Answer:
Find Fx and Fy Fy= F sin 0

Fx= F cos 0 = 30 N sin 30 degree
= 30 N cos 30 degree = 15 N
= 25. 98 N
7. Erica and Tory are out fishing on the lake on a hot summer day when they both decide to go
for a swim. Erica dives of the front of the boat with a force of 45N, while Tory dives off the
back with a force of 60N.
a. Draw a vector diagram of the situation.
b. Find the resultant force on the boat
Answer:
A. 60N 40 N B. R= 60n- 40n = 20 n
8. Ralph is moving the back yard with a push over that he pushes downward with a force of 20.0
N at an angle of 30.0 degree to the horizontal. What are the horizontal and vertical
components of the force exerted by Ralph?
Answer:
Fx= F cos 0
= 20N cos 30 degree
= 17.32 N
Fy= F sin 0
= 20 N sin 30 degree
= 10 N
9. In the record books, there are men who claim that they have such strong teeth that they can
even use them to move cars, trains and helicopters. Joe ponder of love Valley, North Carolina
in one such man. Suppose a car pulling forward with a force of 20,000 N was pulled back a
rope that Joe held in his teeth. Joe pulled the car with a force of 25,000 N. Draw a vector
diagram of the situation and find the resultant force.
Answer:
25,000 __________________ car _________________ 20,000 N

R= 25,000 N- 20,000 n
= 5,000 N
10.Dwight pulls his sister in her wagon with a force of 65N at an angle of 50.0 DEGREE TO THE
VERTICAL. What are the horizontal and vertical components of the fore exerted by Dwight?
Answer:
Find Fx and Fy
Solution Fx= F cos 0
= 65 N cos 40 degree
= 49.79 N
Fy= F sin 0
= 65 N sin 40 degree
= 41.78 N
OPEN UNIVERSITY
CHERRYL R. TOBIAS
MODULE 2
SELF TEST 2
Answer the following:

1. What happens to motion when friction is reduced?
Answer:
If the force is reduced, the motion will continue.
2. Under what condition can a moving object be in a state of equilibrium?
Answer:
A moving object is said to be in a state of equilibrium if it is in constant motion.
3. What must be done to overcome the inertia of an object?
Answer:
Apply a greater amount of external force
4. Is there more friction between two rough objects or two smooth objects?
Answer:
There is more friction between two rough objects
5. Does a state of equilibrium or at rest exist in a tag of war in which one team pulls with a force of
150 N and the other team pulls with the same force in the opposite direction ?how can you tell?
Answer:
Yes, because the amount of forces exerted is equal, therefore the net force is equal to zero, that
means the forces acting on it are balanced.
OPEN UNIVERSITY
CHERRYL R. TOBIAS
MODULE 2
SELF TEST 3
1. What happens when the net force on an object is not zero?
Answer:
The motion and shape of the body will either changed.
2. What happens to the force when either mass or acceleration is increased?
Answer:
If mass or acceleration increased the force is also increased.
3. If you double the net force on an object and keep the mass constant, how will its acceleration
be affected?
Answer:
The acceleration will also double.
4. What will happen to the acceleration if the mass of an object is tripled but the fore is kept the
same?
Answer:
The acceleration will be decreased.
5. Which of the following would require more force to accelerate the same amount:
Answer:
a. Truck b. car c. golf ball
6. The brakers are slammed on a speeding truck and it skids to a stop. If the truck were heavily
loaded so that it had twice the total mass, the skidding distance would be.
Answer:
a. The same
7. Which encounters the greater force of air resistance- a falling elephant or a falling feather?
Explain.
Answer:
The greater force encounter is the elephant because the mass of elephant is greater than the
feather.
8. Which feels a greater pull due to the earth’s gravity, a 10 kg stone or a 20kg stone? If you
drop them, why does the 20kg stone not fall with twice the acceleration of the 10kg stone?
Explain.
Answer:
Greater gravitational force is the 20kg stone, the gravitational force is equal to the air
resistance force, and therefore the net force is zero. If the net force is zero, the acceleration is also
zero and the velocity is constant. Mass will not affect velocity of the object because the air resistance
force is directly opposite to the gravitational force to make a balance forces on the object.
9. A 1kg rock is thrown at 10 m/s straight upward, Neglecting air resistance, what is the net
force that acts on it when it is half way to the top of its path?
Answer:
V= 10m/s
1kg= mass
10.Rose is sledding down an ice covered hill inclined at an angle 15 degree with the horizontal. If
rose and the sled have a combined mass of 54.0 kg, what is the force pulling them down the
hill?
Answer:
F= ma F= 54(9.8)
= 529.8 N
Force pulling them down: Let say friction is Zero
Fo= F sin 0
= 529. 8 N sin 15 degree
= 137.12 N
11. Fellicia, the ballet dancer has a mass of 45.0 kg.

a. What is Felicias weight on earth?
b. What is felicias mass on Jupiter where the acceleration due to gravity is 25.0 m/s22
c. What is Felicias weight on Jupiter?
Answer:
a.M= 45.0 kg
W= mg = 45(9.8) = 441 N
b. 45kg.
c. G= 25m/s2
W= mg= 45(25) = 1,125 N
12.A 20g sparrow flying toward a bird feeder mistakes the pane of glass in a window for an
opening and slams into it with a force of 2.0 N. What is the bird’s acceleration?
Answer:
A= f/m
A=2.0 N/ 20g (1000)
A= 100 m/s2
13.A 30.0 g arrow is shot by William Tell through an 8.00 cm thick apple sitting on top of his sons
head. If the arrow enters the apple at 30.0 m/s and emerges at 25.0 m/s in the same
direction, with the force has the apple resisted the arrow?
Answer:
Vo= 25 m/s Vf= 30 m/s
Vf2= Vo2 + 2 nd Vf=
Vo + at
30(2)= 25 2nd + 2nd 30=25 + at
275=2nd t= 5/a sub.
D= 275/2a d= 25(5/a) + ½ a (5/a)2

D= do +Vot+ ½ at 2 d= 125/a + 25/a
D=1st + ½ AT 2 d= 1075/a
137.5/a = 275/2A A= 2m/s2
F= 30(2) = 60 N
14.Carter’s favorite ride a play land amusement park is the rollercoaster. The rollercoaster car and
passengers have a combined mass of 1620 kg, and they descend the first hill at an angle of
45. O degree to the horizontal. With what force is the rollercoaster pulled down the hill?
Answer:
M= 1620 kg Fp= F sin 0
F= ma= 1620 kg (9.8) = 15876 sin 45 degree
= 15876 N = 11, 226.03 N
15.Flip, an exhausted gymnast, hangs from a bar by both arms in an effort to catch his breath. It
flip has a mass of 65.0 kg, what is the tension in each of Flip arms as he hangs in place?
Answer:
F= Ma m= 65 kg
F= 65 (9.8) = 637 N
2T= F
2T= 637 N
T= 318.5
OPEN UNIVERSITY
CHERRYL R. TOBIAS
MODULE 2
SELF TEST 4
1. What do scientist mean when they say that action equals reaction?
ANSWER
It means that whenever a force is exerted on the second object (action), the second object also
exerts an equal and opposite reaction to the first object (reaction). The magnitude of the reaction
force is equal to the magnitude of the action force but on opposite direction.
2. Give the tree examples of Newton’s Third Law.
ANSWER
Example of Newton’s Third Law of Motion

1. If a table is pressed from its corner with your hand, the table pushes back and result is a
dent in our skin. If we press the table harder, dent in our skin goes deeper.
2. A person slams a wall. Suppose a person slams the wall with his fist. The force applied by
the person through his fist on the wall is equal to the force applied by the wall on the fist. Harder the
person slams the wall, more he gets hurt. Similar procedure is followed when a person kicks the
football.
3. The rock is touched with the toe? When we touch the rock with toe and apply force on it,
the toe exerts force on the rock. Equal amount of force is exerted back on the toe by the rock. If you
apply more force on the rock by the toe, more force is felt by you and it hurts.
3. Newton’s Third Law says that action and reaction are always equal and opposite. If this is
true, why don’t they always cancel one another and leave no unbalanced force acting on any
body?
ANSWER
Because they will act on opposite direction.
4. For every action, there exists an equal and opposite force. Consider action and reaction forces
in the case of a rock falling under the influence of gravity, if action is considered to be that of
the earth pulling down on the rock, can you clearly identify the reaction force?
ANSWER
Gravitational pull is accelerating the rock towards the earth and vice versa, at the same time
acceleration is inversely proportional to its mass. The mass of the rock is very small than the mas of
the earth.
5. If a Mack truck and a Volkswagen have a head- on collision, which vehicle will experience the
greater impact force? Explain?
ANSWER
The Volkswagen have a great impact force, the mass of the Mack truck is more than a
Volkswagen, the 2nd law of motion said that the more mass the more force act on the body.
6. Consider an apple at rest on the table, If we call the gravitational force exerted on the apple
action, what is the reaction force according to Newton’s 3RD Law?
ANSWER
The reaction force is the apply force of the table in the opposite direction of weight of apple.
7-9 incomplete
9. Butch, the 72 kg star quarter back of Belmont high school’s football team, collides with trask, a
stationary left tackle, and is brought to a stop with an acceleration 0f -20 m/s 2.
ANSWER
Butch m= 72 kg a= -20 m/s2
Fa= ma (active force) FA= FR (Reaction Force)

= 72 kg (-20 m/s2) FR= 144 N (Force trask exert on butch)
= 144 N (force butch exert on trask)
10. While chopping down his father’s cherry tree, George discovered that if he swung the axe with a
speed of 25 m/s, it would embed itself 2.3 cm into the tree before coming to a stop.
ANSWER
a. If the axe head had a mass of 2.5 kg, how much force was the three exerting on the axe head
upon impact?
b. How much force did the axe exert back on the three?
ANSWER
AXE m= 2.5 kg v= 25 m/s x= 2.3 cm/ 0.023 m
Vx2= Vxo 2+ 2a (x-xo)

0= 25(2) m/ + 2 a( 0.023)
-65= 0.046 a
A= -13, 586.96 M/S2 Decelerating
FA= FR
a. Reaction Force (FR)
FR=MA
= 2.5 KG (13, 586.96)
=33,967.4 N
b. Active Force ( FA)

FA= FR
FA= 33,967. 4 N
OPEN UNIVERSITY
CHERRYL R. TOBIAS
MODULE 2
Assignment:
1. When a car stops suddenly, the passengers tend to move forward relative to their seats. Why?
When a car makes a sharp turn, the passengers tend to slide to one side of the car. Why?
ANSWER
Because for every action force, there is an equal and opposite reaction force.
3. Why can it hurt your foot more to kick a big rock than a small pebble? Must the big rock hurt
more? Explain.
ANSWER
Because the mass of rock is greater than a small pebble, the greater the mass the more force exert
of the object, at the 3rd law every action there is equal and has an opposite force, we know that the
force exert by foot is equal to the reaction force of big rock. Foot has a less mass than a big rock so
the reaction force is more than the force exerted by the foot.
4. Gunter the weightlifter can lift a 230.0 kg barbell overhead on earth. The acceleration due to
gravity on the sun is 274 m/s2.
a. Would the barbells be heavier on the sun is 274 m/s2.
ANSWER
More heavier on sun because of the acceleration due to the gravity, and also the sun is higher
than the earth.
5. Sammy Sosa swings at a 0.15 kg baseball and accelerate it at a rate of 3.0 x 10(4) m/s2. How
much force does sosa exert on the ball.
ANSWER
M= 0.15 kg a= 3x10(4) m/s2 f=?
F=ma
= 0.15 (3x10(4) m/s2
= 4500 N
6. What is the minimal force a mother must exert to lift her 5.0 kg baby out of its crib?
ANSWER
M= 5 kg g= 9.8 m/s2
W= mg
= 5 kg (9.8)
= 49 N
Minimal force is greater than the weight of baby = 50N
7. Claudia stubs her toe on the coffee table with a force of 100 N.
a. What is the acceleration of Claudias 1.80 kg foot?

b. What is the acceleration of the table if it has a mass of 20.0 kg?
c. Why would Claudia’s toe hurt less if the table had less mass?
ANSWER
FA= 100 N FA= ma m=1.8 kg

A=?
Fa= ma
100N = 1.8 kg (a)
A= 55.56 m/s2
B) F= ma m= 20kg
100N= 20kg (a)
A= 5 m/s2
B. Because mass is directly proportional to the force.
8. After returning home the beach, Samantha hangs get wet 0.20 kg bathing suit in the center of
the 6.0 m long clothesline to dry. This causes the clothesline to sag 4.0 cm. What is the
tension in the clothesline?
ANSWER
M= 20kg w= 20 (9.8) =196 N
Tan 0 = 3/0.04 sin 0 = 0.04/3 cos 0 = 3/3
Fy= Ty + Ty-W
= 2Ty- W W=196 N
0=2(T 0.04/3) – W T= 7350n
W=0.026 T 2t= 14700 N
196= 0.026 T
9. While moving out of her dorm room, Bridget carries a 12kg box to her car, holding it in both
arms.
a. How much force must be exerted by each of her arms to support the box?
b. How will this force change if Bridget holds the box with only one arm?
ANSWER
W= mg = 12 (9.8) = 11.7.6 N
Fy= F+F- W
0= 2f- w
W= 2f
117.6= 2F
A. f= 58.8 N (each arm)
b. only one arm
w=f (one arm)
f= 117.6N
10.Yvette hangs 2.4 kg bird feeder in the middle of a rope tied between two trees. The feeder
creates a tension of 480 N in each side of the rope.
A. If the two trees are 4. 0 m apart, how much will the rope sag in the center?
B. If a bird lands on the feeder, will this have any effect on the tension in the rope? Explain.
ANSWER
W= 2.4(9.8) = 23.52 N
Fy= ty+ty- w
W= 2ty
23.52 n = 2ty
Ty= 11.76 n
Tan 0 = tx/ty
1 = tan -1 479.86 N/11.76 N
= 88.6 degree
Tan 0 = adjacent side/ opposite side
Tan 88.6 degree = 2m/y
Y= 0.022 m/ 2.2 cm
b. Yes because the bird has a mas or weight, the tension on the rope will be increase
because the additional weight cause by bird.
OPEN UNIVERSITY
MODULE 3
LESSON 1. TEMPERATURE, HEAT AND EXPANSION
LESSON 2. TRANSFER OF HEAT
LESSON 3. THERMODYNAMICS
OPEN UNIVERSITY
CHERRYL R. TOBIAS
MODULE 3
SELF TEST 1
ANSWER:
A.
1. Kinetic
2. Temperature
3. Specific heat capacity
4. Calorie
5. fusion
6. vaporization
7. evaporation
8. condensation
B.
1. What is the common way of measuring temperature? Explain.

ANSWER:
By using a thermometer. It is done by immersing the thermometer in the substance (for liquids) or
by placing the thermometer into an intimate contact with the body (for solids)
2. In terms of difference in temperature between objects in thermal contact, in what direction

does heat flows.
ANSWER:
The heat flow from the object with a higher temperature to the object with lower temperature.
3. Why it is not correct to say that matter contains heat?

ANSWER:
Because heat does not contain into the substance. Heat occurs because of the difference in
temperature of the interacting bodies. Once the energy is transferred, it ceases to be heat, it become
internal energy.
4. How is temperature related to the average kinetic energy of molecules that make up a
substance?
ANSWER:
Temperature is the increase of the average kinetic energy of the molecules of the substance. The
higher the temperature, the greater the amount of kinetic energy of the molecules in a substance.
5. What are the effects of adding heat to a substance?

ANSWER:
The effects of adding heat to a substance.
 Thermal expansion
 Phase change (changing from solid to liquid, changing from liquid to gas)
6. How can heating causes a substance to expand?
ANSWER:
By heating, the atoms or molecules gain kinetic energy and they move farther apart that
causes the substance to expand.
7. What are the factors that affect the quantity of heat needed to raise the temperature of a
substance? Explain.
ANSWER:
Factors which affect the amount of heat needed to raise the temperature of a substance are:
 The amount (mass) of the substance – the greater the mass the greater the amount of heat
needed.
 What temperature change is required (from what temperature to what temperature)
 The specific heat of the substance, which is "the number of calories required to raise the
temperature of 1 gram of the substance by 1 degree Centigrade"
 The ambient temperature of the surroundings where this change is to be carried out
8-10
11. The temperature of background radiation left over from the Big Bang during the creation of the universe
is 3K, what is the temperature of the universe
a) In degree C? B) in degree F?
ANSWER:
3 Kelvin
A. In degree C
B. In degree F
Solution:
a. K= C +273 B. F= 1.8 C +32
3= C +273 =1.8 (-270) + 32
C= -273 +3 F= -454
C= -270
12. Peter is heating water on the stove to boil eggs for a picnic. How much heat is required to raise the
temperature of his 10.0 kg vat of water from 20.0 degree Celsius to 100. 0 degree Celsius?
ANSWER:
Heat Energy= (mass) (specific heat capacity) (change in temperature)

= (10kg) (4200 j/kg C) (80 C)
= 3,360, 000 j
13. Nova, who’s mass is 50.0 kg, stays out skiing for too long and her body temperature drop by 2.00 degree
Celsius. What is the amount of heat lost from Nova’s.
ANSWER:
Heat Energy= (50kg) (3470 j/kg C) (2 C)

= 347, 000 j
14. Phoebes insulated foam cup is filled with 0.150 kg of the coffee, that is to hot to drink, so she adds 0.010
kg of milk at 5.0 degree C, if the coffee has an initial temperature of 70.0 degree c and the specific heat
of milk is 3800 j/kg degree C, how hot is the coffee after milk is added? (Assume that no heat leaks out
through the cup.)
ANSWER:
Heat energy of coffee = heat energy of milk
= (0.150) (4200) (Tf- 70 degree C)
= 0.90 (3800) (Tf-5)
= (6300Tf- 44100) = 380 tf – 190
= 630 Tf+ 44,100 = 38tf-190
= -668 tf= -44,290
Tf= 66. 30 degree C
15. Emily is testing her baby’s bath water and finds that is too cold, so she adds some hot water from a
kettle on the stove. If Emely adds 2.00 kg of water at 80.0 degree C to 20.0 kg of bath water at 27.0
degree C, What is the final temperature of the bath water?
ANSWER:
-(2) (4200) (tf-80)
= (20) (4200) (tf-27 degree C)
= -2TF +160 = 20 TF-540
= -22 TF= -700
TF= 31.82 degree C
OPEN UNIVERSITY
CHERRYL R. TOBIAS
MODULE 3
SELF TEST 2
Answer the ff. questions.
1. How is radiation different from conduction and convection? Explain.
ANSWER:
Radiation is different from conduction and convection because in radiation heat transfer occurs
in the absence of a medium to transfer while in conduction and convection heat transfer happens by
collision, direct contact or actual movement of the heated materials.
2. Cite at least 3 situations each of different methods of heat transfer.
ANSWER:
CONDUCTION
Cooking food on the stove is an example of conduction happening

twice – the heat from the burner passes into the metal pan, and then the heat from the metal pan
passes into the food, heating it up.
Touching a hot iron is an example of conduction – the heat passes out of the iron and into
your hand.
Holding an ice cube – the heat is conducted out of your hand, and into the ice cube
(that’s why your hand feels cold)
CONVECTION
Heated up soup for dinner

Warming of the ocean water at daytime.
Using heated air to make a hot-air balloon rise up into the sky
RADIATION
Light bulbs radiate heat
Drying of clothes under the sun
Heating up your body in a bon fire
3. How do wavelengths of radiant energy vary with the temperature of the radiating source?
ANSWER:
The longer the wavelength, the lower the temperature, the shorter the wavelength the higher the
temperature.
OPEN UNIVERSITY
CHERRYL R. TOBIAS
MODULE 3
SELF TEST 3
1. What is the absolute temperature?

ANSWER:
Absolute temperature is lowest possible temperature that a substance may have.
2. Give at least 3 examples of systems which observes the first law of thermodynamics.
ANSWER:
1. Burning a log
2. bicycle wheel will eventually stop spinning unless you push it.
3. kicking a ball
3. How is the first law different from the second law of thermodynamics?
ANSWER:
Difference between the first law and the second law of thermodynamics
Basic idea:
First law: First law of thermodynamics is a version of the law of conservation of energy.

Second law: Second law of thermodynamics states what types of thermodynamic processes are
forbidden in nature.
Content:
First law: First law of thermodynamics states that energy can be neither created nor destroyed.
Second law: It is impossible to construct a perfect heat engine or a perfect refrigerator. It is

impossible to construct a perpetual motion machine. It is impossible to completely convert heat into
work. Heat doesn’t spontaneously flow from a cold reservoir to a hot reservoir. The entropy of an
isolated system never decreases.
Uses:
First law: The equation; ΔU= ΔQ+ ΔW can be used to calculate the algebraic value of one quantity if
other two quantities of the equation are known.
Second law: The second law can be used to calculate the maximum achievable thermal efficiency
(Carnot efficiency) of a given heat engine.
4. How does a heat engine do work?
ANSWER:
Heat engine does work by changing heat into energy in motion.
5. Differentiate internal from external combustion engines.
ANSWER:
The main difference between internal and external combustion engine is that in internal

combustion engines, the working fluid burns inside the cylinder, whereas in external combustion
engines, combustion takes place outside the cylinder and heat is then transferred to the working
fluid.
OPEN UNIVERSITY
MODULE 4
LESSON 1. MAGNETIC AND ELECTRIC FORCES
LESSON 2. ELECTRICITY
LESSON 3. ELECTROMAGNETISM
OPEN UNIVERSITY
CHERRYL R. TOBIAS
MODULE 4
SELF TEST 1
Self-Test-Magnetic and Electric Forces
ANSWER:
1. B (magnetic poles)
2. C (electrostatic induction)
3. A (attracts lightning?
4. A (magnetic)
5. C (an electroscope)
6. B (magnetic domains)
7. B (static electricity)
1. What is magnetism?
ANSWER:
Magnetism is forced produced by electric charges in motion.
2. Why do some matter magnetic?
ANSWER:
Some matter is magnetic because they have magnetic domains.
3. What happens when you bring the north poles of two magnets together? The south poles?
The north pole of one magnet and the south pole of another magnet?
ANSWER:
When you bring the north poles of two magnets together they will repel.
When you bring the south poles of two magnets together they will repel
When you bring the north pole of one magnet to the south pole of the other magnet, they will
attract each other.
4. How is the earth like a magnet?
ANSWER:
The earth is like a giant magnet because it has a magnetic field.
5. What are the two ways that magnetism and electricity are alike?
ANSWER:
Magnetism and electricity are alike in terms of the following:

a. opposite charges attract and like charges repel
b. the force decreases as the magnets/ electric charges are moved farther apart.
6. How magnetism and electricity different?
ANSWER:
In electricity, an object is either positive or negative, while in magnets, there is always be two
poles-north and south. The poles in magnets cannot be separated in two different objects.
7. What is magnetic induction?
ANSWER:
Magnetic induction is the process by which objects become a magnet.
8. How does an electroscope detect and electric charge?
ANSWER:
When a charged object touches the knob, the leaves will separate, therefore it will tell you
that the object is charged. But if not, the knob will not separate.
9. What is ground conductor?
ANSWER:
Grounded conductor cannot hold a charge because any charge that passes through it flows
into the electrical ground.
10.Why do clothes sometimes cling together when they came out of the dryer?
ANSWER:
Clothes stick together in the dryer due to static electricity. Different fabrics exchange electrons as
they rub against each other in the dryer. Some clothes become positively charged and others become
negatively charged. These opposite static charges cause the clothes to stick together and produce
crackling sparks when pulled apart.
Lesson 2: Electricity
ANSWER:
1. The light in simple circuit is brighter than the light in the series circuit.
2. The brightness of lights in the parallel circuit is the same in the simple circuit.
3. The other bulb will not light.
4. The other bulb will be lighted.
5. The dimness is due to the voltage drop in each circuit. The current in each circuit is equal but
the total voltage is distributed in the different bulbs.
6. When one light bulb goes out, all the light bulb in the series circuit go out because the flow of
current is being stopped ( open circuit) at the point where the bulb were goes out.
OPEN UNIVERSITY
CHERRYL R. TOBIAS
MODULE 4
SELF TEST 2
A. ANSWERS:
1. Electricity
2. true
3. positive terminal
4. true
5. short
6. series circuit
7. voltage
8. alternating current
9. true
10. circuit breaker
B. ANSWER THE FF. QUESTIONS
ANSWERS:
1. Electric current is the flow of electric charges from one point to another.
2. Voltage moves electric current
3. ampere- current, volt-voltage, ohm- resistance, watt- electric power
4. Circuit is complete path that allows the flow of current.
5. Short circuit is dangerous because it can cause fire.

6. In a series connection, there is single path for a current to flow while in parallel connection,
the current floes or divides into several or separate paths.
7. The electric outlet is like the terminals of a battery is such a way that he outlet and the
terminal of the battery allows the flow of current from the source to the load and vice versa.
8. Ohms law gives us the idea on the relationship between current, voltage and resistance. It
states that current is directly proportional to the voltage but inversely proportional to its
resistance.
9. Voltage is created in a flashlight battery by the chemical reactions between a moist paste and
a metal.
10.In direct current, current always flows in the same direction while in alternating current,
current flows in constantly switching directions.
Lesson No. 3: Electromagnetism
ANSWERS:
How was the concept of electromagnetism discovered?
The concept of magnetism was discovered by Han’s Christian Oersted by placing a compass
just above a wire carrying an electric current. To his surprise, the compass needle –small magnet-
moved from a position pointing north to a position at right angle to the wire.When he reversed the
direction of the current, the needle reversed direction. But it was still at a right angle to the wire.
When the current turn off, the needle pointed north again.
What happens when you put a metal bar that can be permanently magnetized inside a coil?
The bar will be magnetized by the magnetic field. When the current is turn off, the bar will
remain magnetized.
Now, what happens when the bar that you placed inside the coil is made up of iron that loses its
magnetism easily?
When the current flows through the coil, the bar will be magnetized. But as soon as you turn
off the current, the bar will lose its magnetism. When you change the direction of the current, the
bar is magnetized in opposite direction.
OPEN UNIVERSITY
CHERRYL R. TOBIAS
MODULE 4
SELF TEST 3
ANSWER THE FF. QUESTIONS.
1. How are the electromagnets made?

ANSWERS:
Electromagnets are made from magnets in which its magnetism is produced by an electric
current.
2. How is electric current produce by electromagnetic induction?

ANSWERS:
Electric current is produced by electromagnetic induction by changing magnetic field.
3. How is electricity produced by an electric generator?
ANSWERS:
Electricity is produced by an electric generator by electromagnetic induction to change kinetic

energy into electrical energy.

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Republic of the Philippines

UNIVERSITY OF NORTHERN PHILIPPINES

DLS 212 – SELECTED TOPICS

DLS 212 – Selected Topics in General Physics

LESSON 1. LINEAR MOTION

LESSON 2. NONLINEAR MOTION

DLS 212 – Selected Topics in General Physics

Answer the following questions:

B. Velocity- speed with direction.

C. Acceleration- change of velocity over a period of time.

Distance from start Time(s)

 20m t= t2- t1 40-2=38 sec.

 10m v= 10m/ 2 sec = 5m/sec

4. From the graphs given below, describe the acceleration of a particle.

Time (Total) = Time travel from A To B + Time travel from B to A

Time (Total) = Time travel from A to B + Time travel from B to A

6. What is the average speed of the Calesa (Refer to prob.3)

DLS 212 – Selected Topics in General Physics

DLS 212 – Selected Topics in General Physics

SELF TEST III

2. At what part of its trajectory does a projectile have

a. minimum speed- when the projectile is at lowest

b. maximum speed- when the projectile is at highest

3. Discuss the difference between linear speed and rotational speed.

Given: g= 10 m/s2 r= 10 m Find: V=?

14.Roxanne is making a strawberry milkshake in her blender. A tiny, 0.0050 kg strawberry is

DLS 212 – Selected Topics in General Physics

1. What is time rate of change? Give three examples.

2. Distinguish between speed and velocity.

Y= Yo + V y o t – ½ g t 2 to the ground. Because as he start, the velocity at ground is decelerating

T= 1.12 sec reached the ground.

Location: From Center Jessica r= 8m Julie r= 6 m

W= 1 rotation/ 40 sec A. Jessica has the greater linear speed

Republic of the Philippines

DLS 212 – Selected Topics in General Physics

LESSON 2. FIRST LAW OF MOTION

LESSON 3. SECOND LAW OF MOTION

LESSON 4. THIRD LAW OF MOTION

Republic of the Philippines

DLS 212 – Selected Topics in General Physics

A. MODIFIED TRUE OR FALSE.

B. Give what are asked in the ff.

2. Explain the difference between mass and weight.

Find Fx and Fy Fy= F sin 0

25,000 __________________ car _________________ 20,000 N

DLS 212 – Selected Topics in General Physics

Answer the following:

2. Under what condition can a moving object be in a state of equilibrium?

3. What must be done to overcome the inertia of an object?

There is more friction between two rough objects

DLS 212 – Selected Topics in General Physics

1. What happens when the net force on an object is not zero?

2. What happens to the force when either mass or acceleration is increased?

11. Fellicia, the ballet dancer has a mass of 45.0 kg.

275=2nd t= 5/a sub.

D= 275/2a d= 25(5/a) + ½ a (5/a)2

DLS 212 – Selected Topics in General Physics

Answer the following:

25,000 __ car _ 20,000 N