Here is the Final UPCAT Sample Quiz in the Science Subtest. Remember that we do not have any timers for the quizzes so you may want to time yourself here. If you find any errors in the quiz, please dont hesitate to let us know about it through the comments section. Moreover, if you have a better solution set or explanation that is provided in the solution sets, please dont hesitate to share it too.
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Physics Quiz
Question 1

F = (200)(cos 20^{o})


F=200/sin 20^{o}


F=200/cos 20^{o}


F = (200)(sin 20^{o})

Sin20 = 200/F
sin20 X F = 200
F = 200/sin20
Question 2

1.50 kJ


2.00 kJ


2.25 kJ


1.75 kJ

m = 500 kg v = 3.0 m/s
KE = 1/2 (500kg) (9m^2/s^2) = 2.25 KJ
Question 3

Polarization


optical illusion


reflection


refraction

Question 4

4 m/s^{2}


3 m/s^{2}


1 m/s^{2}


0.25 m/s^{2}

Question 5

displacement


force


speed


acceleration

Question 6

A car running along a curve


A cart moving downhill on a mountain slope


A ball thrown upwards


A book on the table

Question 7

The mirror was concave.


The mirror was convex.


The mirror has nothing to do with it.


The mirror was taller.

Question 8

D


B


A


C

Question 9

V_{f} = V_{i}^{2} + 2aΔt


V_{f} = V_{i} + aΔt


V_{f} = 2V_{i} + aΔt


V_{f} = V_{i} – aΔt

Question 10

600 N


400 N


500 N


300 N

w= 150N therefore for the block to remain stable, the frictional force should also be 150N..
using the equation u = Ff/Fn
0.25 = 150 N / Fn
Fn = 150N/0.25
Fn = 600N
since the normal force is an equal but opposite force to the one pushing the block to the wall, the final answer is also 600N.
Question 11

D


B


A


C

Question 12

They require a medium for transmission


They are transverse waves.


They travel with the same speed in the absence of vacuum.


They are produced by moving charges.

Question 13

Frequency


Wave length


Pitch


Amplitude

Question 14

A ball falling from an airplane.


A tanker in full speed turning its engine in reverse an hour before reaching the port.


A train slowing down before reaching the station.


A man on a bus being “pushed” backward when the bus starts moving.

Question 15

kg∙m/s


kg/L


kg∙m^2


kg∙m/s^2

Question 16

The force of the ball against the bat.


The grip of the player’s hand on the bat.


The air resistance on the ball.


Muscular effort in the player’s arm.

Question 17

4.5 KN


3.0 KN


1.5 KN


2.5 KN

force = [(100 kg)(900m^2/s^2)/60m = 1500kgm/s^2
= 1500 N = 1.5 kN
Question 18

40 cm


20 cm


80 cm


60 cm

\(\frac{L_{1}}{F_{1}}=\frac{L_{2}}{F_{2}}\)
\(L_{2}=\frac{L_{1}F_{2}}{F_{1}}= \frac{20cmX1500N}{500N}= 60cm\)
Question 19

ordinary speed


average speed


instantaneous speed


uniform speed

Question 20

Could not be determined.


They stick together.


They bounce back with the same speed and distance.


They bounce sideways.

Question 21

2 N


1 N


3 N


4 N

Given : m = 2 kg a = 0.5 m/s^{2}
F = mass x acceleration
F = 2 kg x 0.5 m/s^{2} = 1 kg m/s^{2} = 1 N
Question 22

Because the fish moves too fast.


Because light bounces off the ocean floor.


Because light travels at a different speed on water.


This phenomenon is not true.

Question 23

ultrasonic


mega sonic


infrasonic


supersonic

Question 24

1.1 Km


1.3 Km


0.7 Km


0.9 Km

a = 0.5 km
b= 1.2 km
c = ?
c = \(\sqrt{a^{2}+b^{2}}\)
c = 1.3
Question 25

III only


I and II only


II only


I only

Question 26

kinetics


Mechanics


kinematics


Static

Question 27

Wavelength


Frequency


penetrability


Energy

Question 28

Sound is equal the speed of light.


Light is 17860 mph faster than sound.


Sound is 740 mph faster than light.


Light is 740 mph faster than sound.

Question 29

rectilinear motion


uniform circular motion


horizontal motion


projectile motion

Question 30

KE < PE


KE = PE


KE > PE


KE = PE

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kailan po yung UPCAT for batch 20122013?
The UPCAT for AY 20122013 freshmen was held in August 2011. The UPCAT for AY 20132014 freshmen (those who want to become UP Freshies in June 2013) will be administered in August 2012.
i need to practice more ) i only got 12 )
At least you still have time to practice. Incoming 4th year HS student?
@Arvi, Is that how it is? But I thought the heavier person should be much closed to the center so that the seesaw would balance. Since the distance of the lighter person is 20cm from the center, shouldn’t the distance of the heavier person be smaller?
kelan po deadline ng application form ng ustet
Hi!
I’m Arvi from UP Diliman, Eng’g Batch ’03.
I’d like to correct the solution to question no. 25. Following your line of explanation, L1/ F1 = L2/ F2, we can’t simply express the moment arms (L1 & L2) proportional to the forces (F1 & F2) in this manner. In Physics, this is governed by statics, wherein the summation of forces and moments must equal to zero at equilibrium (to prevent the seesaw from moving or changing its horizontal position.) Hence, the correct proportion should have been L1/F2 = L2/F1 or L1xF1 (moment of F1) = L2 x F2 (moment of F2) using the fulcrum (center of seesaw) as our reference for the moments. Wherein L1 & L2 are the moment arms, F1 & F2 are the forces or weights of the bodies. Once you get L2, subtract it from 1m or 100cm to get the answer.
Moreover, using the center of the seesaw as our point to get the summation of moments, L1 would be 80cm from the fulcrum. F1 is 500N. The moment of force would be F1 x L1 (500N x 80cm = 40000 N.cm, counterclockwise) This moment must be equal to the moment of force, F2 = 1500N in order to maintain equilibrium (F2 will cause a moment in the clockwise direction). The position, x, from the fulcrum multiplied by F2 must be equal to 40,000 N.cm. Solving for x; 40000 = 1500x, x = 26.67cm. Hence, the answer must be 100 – x = 100 – 26.67 = 73.33 from the right side of the seesaw.
Therefore, the answer in not in the given choices. Just my two cents.