Here is the Final UPCAT Sample Quiz in the Science Subtest. 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

*Physics Quiz*. You scored %%SCORE%% out of %%TOTAL%%. Your performance has been rated as %%RATING%%

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Question 1 |

A | None of the above. |

B | It is accelerating tangent to the circle. |

C | It is accelerating towards the center. |

D | It is not accelerating. |

Question 2 |

A | half |

B | one-third |

C | twice |

D | the same |

Question 3 |

A | 2000 N |

B | 20000 N |

C | -20000 N |

D | -2000 N |

Question 4 |

A | Theta |

B | Reflection |

C | Periods |

D | Cycles |

Question 5 |

A | instantaneous speed |

B | average speed |

C | uniform speed |

D | ordinary speed |

Question 6 |

A | Energy |

B | penetrability |

C | Frequency |

D | Wavelength |

Question 7 |

A | Velocity |

B | Acceleration |

C | Momentum |

D | Power |

Question 8 |

A | 3.2G N |

B | 320G N |

C | 3200G N |

D | 32G N |

Question 9 |

A | Halved |

B | Quartered |

C | Tripled |

D | Doubled |

Question 10 |

A | 25 m/s |

B | 2.14 m/s |

C | 2.5 m/s |

D | 15 m/s |

Question 11 |

A | Momentum |

B | Force |

C | Impulse |

D | Acceleration |

Question 12 |

A | 0.25 m/s ^{2} |

B | 4 m/s ^{2} |

C | 1 m/s ^{2} |

D | 3 m/s ^{2} |

^{2}

Question 13 |

A | The force of the ball against the bat. |

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

C | The air resistance on the ball. |

D | Muscular effort in the player’s arm. |

Question 14 |

A | Gravity is affected by altitude |

B | Acceleration due to gravity is constant |

C | Gravity changes the mass of an object |

D | Acceleration due to gravity is imaginary |

Question 15 |

A | -55 N |

B | None of the above |

C | 12 meters |

D | 34 kph |

Question 16 |

A | They require a medium for transmission |

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

C | They are transverse waves. |

D | They are produced by moving charges. |

Question 17 |

A | 60 cm |

B | 40 cm |

C | 80 cm |

D | 20 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 18 |

A | 4 Hz |

B | 6 Hz |

C | 10 Hz |

D | 2 Hz |

Question 19 |

^{2})

A | 100 m |

B | 110 m |

C | 10 m |

D | 11 m |

Question 20 |

^{2})

A | 5500 W |

B | 2341.12 W |

C | 1567.13 W |

D | 1718.75 W |

Question 21 |

A | t = 4 secs to t = 6 secs. |

B | t = 2 secs to t = 4 secs. |

C | t = 0 secs to t = 2 secs. |

D | t = 6 secs to t = 8 secs. |

Question 22 |

A | C |

B | B |

C | D |

D | A |

Question 23 |

A | 2 m/s |

B | 75 m/s |

C | 93.33 m/s |

D | 36.67 m/s |

Question 24 |

A | A scalar has direction while a vector doesn’t. |

B | A vector is larger than a scalar. |

C | A vector has direction while a scalar doesn’t. |

D | A scalar is a typed of vector. |

Question 25 |

A | Newton |

B | m/s2 |

C | Joules |

D | \(\frac{Kg\cdot m}{s}\) |

Question 26 |

A | The object is moving to the left at constant speed. |

B | The object is at rest. |

C | The object is moving to the right at constant speed. |

D | All of the above. |

Question 27 |

A | Acceleration |

B | Wavelength |

C | Speed |

D | Frequency |

Question 28 |

A | Inertia is directly proportional to speed. |

B | Inertia is a force that keeps objects at rest or relative motion. |

C | Inertia is the force that stops motion. |

D | The bigger the object the more inertia it has. |

Question 29 |

A | A ball falling from an airplane. |

B | A train slowing down before reaching the station. |

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

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

Question 30 |

^{2}?

A | 3 N |

B | 1 N |

C | 2 N |

D | 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

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Number 13

Mali yung computation,

Vf=Vi+at

120=0+a(5)

a=120/5

a=24m/s

so kung force hinahanap F=ma

F=(800kg)(24m/s)

F=19200 N

wala yung sagot sa choices

22/30 🙁 practice pa moreeeee!bring it on!

Correction: L1 = 80cm not L2. L2 = 80cm x 500N / 1500N, L2 = 26.67cm (means that 1500N is 26.67cm from the fulcrum).

Haha tnx d2 mas nging broad yung mind ko sa lalabas sa UPCAT <3

Help me po,Im taking UPCAT this August,give me somd tips,hints sa mga pweding lumabas this year’s UPCAT,09273336362,number ko,text or call me nalang f u can help me. Thanks in advance

Help me po,Im taking UPCAT this August,give me somd tips,hints sa mga pweding lumabas this year’s UPCAT

kailan po yung UPCAT for batch 2012-2013?

The UPCAT for AY 2012-2013 freshmen was held in August 2011. The UPCAT for AY 2013-2014 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?

Pls re-read. It’s 20cm from the left side. So it’s 80cm from the center given the length of seesaw is 2m (200cm), half of it being 1m (100cm). I agree with Arvi.

Hope I helped! 🙂

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 see-saw 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 see-saw) 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 see-saw 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 see-saw.

Therefore, the answer in not in the given choices. Just my two cents.

ang galing galing nman..

i agree on your equation ARVI, which is L1/F2 = L2/F1, but with respect to the given length of the see-saw which is 2meters (200cm), the distance from the fulcrum on the left side is 180cm (L1=180cm). F1 and F2 are the weight of the boy and the man (F1=500N, F2=1,500N). Thus. L2= (L1*F1)/F2,

L2=(180cm*500N) / 1,500N

L2=90,000cmN / 1,500N

L2=60cm

Hence, the answer remains to be 60cm (which is included on the choices) however the explanation should be edited.

Fulcrum is in the middle of the seesaw, therefore L2=80cm not 180cm.