# Phy 111 - Elementary Mechanics

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• 1.

### Assuming that the Earth is a uniform sphere of radius 6.4 x 106 m and mass 6.0 x 1024 kg, find the gravitational field strength g at a point on the surface of the earth

• A.

19.77ms-2

• B.

199.10ms-2

• C.

9.77ms-2

• D.

100.7ms-2

C. 9.77ms-2
Explanation
g = GM / r2 = (6.67 × 10-11)(6.0 x 1024) / (6.4 x 106)2 = 9.77ms-2
where
G = 6.67 × 10-11
M = 6.0 x 1024
r = 6.4 x 106

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• 2.

### Assuming that the Earth is a uniform sphere of radius 6.4 x 106 m and mass 6.0 x 1024 kg, find the gravitational field strength g at a point: at height 0.50 times the radius of above the Earth's surface

• A.

14.314ms-2

• B.

24.134ms-2

• C.

10.314ms-2

• D.

4.34ms-2

D. 4.34ms-2
Explanation
g = GM / r2 = (6.67 × 10-11)(6.0 x 1024) / ( (1.5 × 6.4 x 106)2 = 4.34ms-2

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• 3.

### What is the gravitational field strength at a height h above the surface of the Earth? R is the radius of the earth

• A.

gR^2/(R+h)2(gR2)/(R+h)2

• B.

GR/(R+h)2

• C.

GR/(R+h^2)2

• D.

GR/(R^2+h)2

A. gR^2/(R+h)2(gR2)/(R+h)2
Explanation
The gravitational field strength at a height h above the surface of the Earth is given by gR^2/(R+h)2. This formula takes into account the radius of the Earth (R) and the distance from the surface (h) to calculate the gravitational field strength. The numerator gR^2 represents the gravitational force at the surface of the Earth, and the denominator (R+h)^2 accounts for the increase in distance from the center of the Earth. This equation shows that as the height above the surface increases, the gravitational field strength decreases.

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• 4.

### The instrument used for experimental determination of the value of the universal gravitational constant G is

• A.

Galilioâ€™s telescope

• B.

Ballistic pendulum

• C.

Newtonâ€™s balance

• D.

Cavendish balance

D. Cavendish balance
Explanation
Isaac Newton's law of universal gravitation proposed that the gravitational attraction between any two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. In equation form, this is often expressed as follows:

The constant of proportionality in this equation is G - the universal gravitation constant. The value of G was not experimentally determined until nearly a century later (1798) by Lord Henry Cavendish using a torsion balance.

Cavendish's apparatus for experimentally determining the value of G involved a light, rigid rod about 2-feet long. Two small lead spheres were attached to the ends of the rod and the rod was suspended by a thin wire. When the rod becomes twisted, the torsion of the wire begins to exert a torsional force that is proportional to the angle of rotation of the rod. The more twist of the wire, the more the system pushes backwards to restore itself towards the original position. Cavendish had calibrated his instrument to determine the relationship between the angle of rotation and the amount of torsional force. A diagram of the apparatus is shown below.

or the real diagram

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• 5.

### A passenger in a moving car and a passerby standing at the road side see each other as moving in the opposite direction. Which of the following is not true?

• A.

The passer-by is stationery relative to the passenger

• B.

Both observers are in motion relative to each other

• C.

The passenger is in motion relative to the passer-by

• D.

The passer-by is in motion relative to the passenger

B. Both observers are in motion relative to each other
Explanation
Both observers are in motion relative to each other. This is not true because according to the question, the passenger in the car and the passerby see each other as moving in the opposite direction. This implies that they are moving relative to each other, but not both in motion relative to each other.

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• 6.

### Which of the following statements is not correct about reference frames?

• A.

Any reference frame moving at constant velocity with respect to an inerial refernce frame is also inertial

• B.

In non-inertial reference frames the motion of objects depend only on the interactions of constituent particles among themselves

• C.

If the coordinate axes of a reference frame attached to an object remain fixed in space then the object is at rest

• D.

Laws of physics are invariant in inertial reference frames

D. Laws of physics are invariant in inertial reference frames
Explanation
The statement "Laws of physics are invariant in inertial reference frames" is correct. Inertial reference frames are frames of reference in which Newton's first law of motion holds true, meaning that an object at rest will remain at rest and an object in motion will continue to move with constant velocity, unless acted upon by an external force. In these frames, the laws of physics, including the conservation of momentum and energy, are invariant, meaning they hold true regardless of the reference frame. This principle is a fundamental concept in classical mechanics.

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• 7.

### What are the dimensions of power (time rate of change of expending energy)

• A.

MLT−2

• B.

ML2Tâˆ’3

• C.

ML2Tâˆ’2

• D.

M2L2Tâˆ’3

B. ML2Tâˆ’3
Explanation
The dimensions of power, which is the time rate of change of expending energy, are given by ML2Tâˆ’3. This means that power is measured in units of mass (M), length (L), and time (T), with the exponents indicating the power to which each dimension is raised. The exponent of -3 for time (T) indicates that time is in the denominator, reflecting the inverse relationship between power and time. The exponents of 2 for mass (M) and length (L) indicate that power is directly proportional to the square of mass and length.

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• 8.

### Two forces act on a point object as follows: 100N at 170derees and 100N at 50degrees. Find their resultant force

• A.

100 N at 110o

• B.

110 N at 50o

• C.

110 N at 100o

• D.

100 N at 50o

A. 100 N at 110o
Explanation
100sin10+ 100sin50 =Vertical component
=93.96 N.
Horizontal=
-100cos10+100cos50=-34.20N.

Resultant = squreoot(93.96)^2+(-34.20)^2
=aprox. 100N. making angle of 110 degress with the positive x -axes

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• 9.

### The speed of 90 hm/hr is equal to ------------------ m/s

• A.

324

• B.

90

• C.

15

• D.

25

D. 25
Explanation
If
1 km = 1000 m

and

1 h = 60 x 60 =3600 s

divide equation 1 by 2

therefore,
1 km/h = 1000m / 3600s

so

90 km/h = (90 x 1000 )m
3600 s
= 25 m/s

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• 10.

### Type question here

• A.

• B.

• C.

• D.

A. Answer option 1
• 11.

### Identify the instrument above, which always spin continuously when torqued?

• A.

Horoscope

• B.

Stetoscope

• C.

Gyroscope

• D.

Spinoscope

C. Gyroscope
Explanation
A gyroscope is a device for measuring or maintaining orientation, based on the principles of angular momentum.[1] Mechanically, a gyroscope is a spinning wheel or disc in which the axle is free to assume any orientation. Although this orientation does not remain fixed, it changes in response to an external torque much less and in a different direction than it would without the large angular momentum associated with the disc's high rate of spin and moment of inertia.
The fundamental equation describing the behavior of the gyroscope is

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• 12.

### A monkey in a perch 20m high in a tree drops a coconut above the head of a zoo keeper as he runs with a speed 1.5m/s beneath the tree. How far behind him in metres does the coconut hit the ground?

• A.

3.03

• B.

2.25

• C.

3.06

• D.

2.02

A. 3.03
Explanation
Seeing this problem, the correct approach would be to notice that the zookeeper runs at a constant velocity. So, for our final step, we will use the equation

V = distance/time
U = d / t
but,     d  = ut
u = initial velocity = 1.5m
t = ?
d = ?
wow, so wot do we do?
considering when the coconut is falling, something happen,
s  = ut + ½ g t²
since as it falls, v=unkown then u = 0m/s²,
s  = ut + ½ g t² = s  = (0)t + ½ g t² s  = ½ g t²
where, s = 20m, g = 10m/s²,
20 = ½ (10) t²
40= 10 t²
4=t² , t = 2s
recall, .
U = d / t
but,     d  = ut
u = initial velocity = 1.5m
t = 2s
d = ?
d = 2 (1.5) = 3m
d=3m
Therefore, the zookeeper had run a total of 3m.

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• 13.

### A monkey in a perch 20 m high in a tree drops a coconut above the head of a zoo keeper as he runs with a speed 1.5 m/s beneath the tree actually intending to hit the toes of the zoo keeper, how early in seconds should the coconut be dropped by the monkey

• A.

1.35

• B.

1.50

• C.

2.04

• D.

3.02

C. 2.04
Explanation
Seeing this problem, the correct approach would be to notice that the zookeeper runs at a constant velocity. So, for our final step, we will use the equation

V = distance/time
U = d / t
but,     d  = ut
u = initial velocity = 1.5m
t = ?
d = ?
wow, so wot do we do?
considering when the coconut is falling, something happen,
s  = ut + ½ g t²
since as it falls, v=unkown then u = 0m/s²,
s  = ut + ½ g t² = s  = (0)t + ½ g t² s  = ½ g t²
where, s = 20m, g = 10m/s²,
20 = ½ (10) t²
40= 10 t²
4=t² , t = 2s

Therefore, the zookeeper had run a total of 2m, all these actions happened simultaneously.

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• 14.

### Which one do you like?

• A.

Option 1

• B.

Option 2

• C.

Option 3

• D.

Option 4

A. Option 1
Explanation
The given question asks for personal preference among the options. Since the answer states "Option 1", it implies that the person prefers Option 1 over the other options.

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• Current Version
• Mar 21, 2023
Quiz Edited by
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• Oct 08, 2013
Quiz Created by
Timetconcepts

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