Question 1

You are facing east looking at the clock on the wall.You see the second hand moving clockwise (of course).In what direction is the angular velocity vector for this second hand?
A) east
B) west
C) north or south,but not east or west
D) For half the rotation it is east,and for the other half of the rotation it is west.

Accepted Answer

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Question 2

What angular speed (in revolutions/second)is needed for a centrifuge to produce an acceleration of 1 000 g at a radius arm of 15.0 cm?
A) 40.7 rev/s
B) 75.4 rev/s
C) 81.4 rev/s
D) 151 rev/s

Accepted Answer

The answer of What angular speed (in revolutions/second)is needed for...

Question 3

An object of mass 0.50 kg is transported to the surface of Planet X where the object's weight is measured to be 20 N.The radius of the planet is 4.0 × 10⁶ m.What free fall acceleration will the 0.50-kg object experience when at the surface of Planet X? A) 48 m/s² B) 20 m/s² C) 16 m/s² D) 40 m/s²

Accepted Answer

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Question 4

A roller coaster,loaded with passengers,has a mass of 2 000 kg;the radius of curvature of the track at the bottom point of the dip is 24 m.If the vehicle has a speed of 18 m/s at this point,what force is exerted on the vehicle by the track? (g = 9.8 m/s²) A) 2.3 × 10⁴ N B) 4.7 × 10⁴ N C) 3.0 × 10⁴ N D) 1.0 × 10⁴ N

Accepted Answer

The force exerted on the roller coaster by the track can be calculated using the formula: 
Centripetal force = (mass x velocity^2) / radius

Plugging in the given values, we get: 
Centripetal force = (2,000 kg x (18 m/s)^2) / 24 m 
Centripetal force = 4,500 N

Therefore, the force exerted on the vehicle by the track is 4.7 x 10^3 N (rounded to two significant figures). The closest answer choice is B.

Question 5

An object of mass 0.50 kg is transported to the surface of Planet X where the object's weight is measured to be 20 N.The radius of the planet is 4.0 × 10⁶ m.What free fall acceleration will the 0.50-kg object experience when transported to a distance of 2.0 × 10⁶ m from the surface of this planet? A) 90 m/s² B) 20 m/s² C) 13 m/s² D) 18 m/s²

Accepted Answer

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Question 6

Consider a child who is swinging.As she reaches the lowest point in her swing: A) the tension in the rope is equal to her weight. B) the tension in the rope is equal to her mass times her acceleration. C) her acceleration is downward at 9.8 m/s². D) none of the above.

Accepted Answer

At the lowest point in a swing, the tension in the rope is greater than the child's weight because it must support the weight and provide the centripetal force necessary for circular motion. The child's acceleration is not downward at this point; instead, the acceleration is centripetal, directed towards the center of the circular path.

Question 7

If a planet has 3 times the radius of the Earth,but has the same density as the Earth,what is the gravitational acceleration at the surface of the planet? (g = 9.8 m/s²) A) 29.4 m/s² B) 88.2 m/s² C) 265 m/s² D) 3.27 m/s²

Accepted Answer

The gravitational acceleration at the surface of a planet is proportional to its radius if the density is constant. Since the planet's radius is 3 times that of Earth, the gravitational acceleration will be 3 times that of Earth, which is 3 * 9.8 m/s² = 29.4 m/s².

Question 8

Consider a point on a bicycle tire that is momentarily in contact with the ground as the bicycle rolls across the ground with constant speed.The direction for the acceleration for this point at that moment is:
A) upward.
B) down toward the ground.
C) forward.
D) at that moment the acceleration is zero.

Accepted Answer

The point on the bicycle tire that is in contact with the ground is momentarily at rest relative to the ground but is about to move upward and forward as the wheel rolls. This change in velocity indicates acceleration, which is directed upward at that moment, as the point moves from the bottom of the wheel to a higher position on the wheel's circumference.

Question 9

The Earth is 93 million miles (mi)from the Sun and its period of revolution is 1 year = 3.15 × 10⁷ s.What is the acceleration of the Earth in its orbit about the Sun? A) 18.6 mi/s² B) 9.3 × 10−³ mi/s² C) 13.6 × 10−⁶ mi/s² D) 3.7 × 10−⁶ mi/s²

Accepted Answer

The acceleration of the Earth in its orbit is given by:

a = (4π^2r) / T^2

where r is the distance between the Earth and the Sun, and T is the period of revolution.

Substituting the given values:

a = (4π^2 * 93,000,000) / (3.15 × 10^7)^2
  = 3.7 × 10^-16 mi/s^2

Converting to scientific notation with proper units:

a = 3.7 × 10^-16 mi/s^2
  = 3.7 × 10^-8 m/s^2
  = 3.7 × 10^-8 / 1.609 km/s^2  (converting miles to kilometers)

a = 2.3 × 10^-8 km/s^2

Therefore, the best choice is (D) 3.7 × 10^-16 mi/s^2.

Question 10

If a planet has a radius 20% greater than that of the Earth but has the same mass as the Earth,what is the acceleration due to gravity at its surface? A) 14 m/s² B) 12 m/s² C) 8.2 m/s² D) 6.8 m/s²

Accepted Answer

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Question 11

An object when orbiting the Earth at a height of three Earth radii from the center of the Earth has a weight of 1.00 N.What is the object's mass? (g at the surface of the Earth is 9.8 m/s²) A) 0.102 kg B) 0.306 kg C) 0.92 kg D) 1.0 kg

Accepted Answer

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Question 12

The acceleration due to gravity at the surface of Planet X is 10 m/s².What is the acceleration due to gravity at an altitude of 3 000 km above the surface of this planet? A) 10 m/s² B) 8.0 m/s² C) 4.4 m/s² D) More information is needed.

Accepted Answer

The acceleration due to gravity at a certain altitude above a planet's surface depends on the mass of the planet and the distance from the planet's center to the point of interest (altitude plus the planet's radius). Without knowing the radius of Planet X, we cannot calculate the new acceleration due to gravity.

Question 13

An airplane in a wide sweeping "outside" loop can create zero gees inside the aircraft cabin.What must be the radius of curvature of the flight path for an aircraft moving at 150 m/s to create a condition of "weightlessness" inside the aircraft?
A) 1 150 m
B) 1 800 m
C) 2 300 m
D) 3 600 m

Accepted Answer

In order to create a condition of weightlessness inside the aircraft, the centrifugal force acting on the aircraft must be equal and opposite to the force of gravity. This can be achieved when the aircraft is in a state of freefall.

When an aircraft is in a wide sweeping "outside" loop, it follows a circular path with a centripetal acceleration given by:

a = v^2 / r

where v is the velocity of the aircraft and r is the radius of curvature of the flight path.

In order to create zero gees, the centripetal acceleration a must be equal and opposite to the acceleration due to gravity g, so:

a = g

v^2 / r = g

r = v^2 / g

Substituting v = 150 m/s and g = 9.81 m/s^2:

r = (150 m/s)^2 / 9.81 m/s^2

r = 2 304 m

Therefore, the radius of curvature of the flight path must be 2 300 m, which corresponds to choice C.

Question 14

A wheel is rotated about a horizontal axle at a constant angular speed.Next it is rotated in the opposite direction with the same angular speed.The acceleration at a point on the top of the wheel in the second case as compared to the acceleration in the first case:
A) is in the same direction.
B) is in the opposite direction.
C) is upward.
D) is tangential to the wheel.

Accepted Answer

The acceleration at a point on the top of the wheel is directed towards the center of the circle and is therefore perpendicular to the surface of the wheel. When the wheel is rotating in one direction (first case), the direction of the acceleration is towards the center of the circle, while in the second case (rotating in the opposite direction), the direction of the acceleration is still towards the center of the circle. Therefore, the direction of the acceleration at the top of the wheel remains the same in both cases.

Question 15

An Earth satellite is orbiting at a distance from the Earth's surface equal to one Earth radius (4 000 miles).At this location,the acceleration due to gravity is what factor times the value of g at the Earth's surface?
A) There is no acceleration since the satellite is in orbit.
B) 2
C) 1/2
D) 1/4

Accepted Answer

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Question 16

At what speed will a car round a 52-m-radius curve,banked at a 45° angle,if no friction is required between the road and tires to prevent the car from slipping? (g = 9.8 m/s²) A) 27 m/s B) 17 m/s C) 23 m/s D) 35 m/s

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Question 17

At what angle (relative to the horizontal)should a curve 52 m in radius be banked if no friction is required to prevent the car from slipping when traveling at 12 m/s? (g = 9.8 m/s²) A) 28° B) 32° C) 16° D) 10°

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Question 18

A cylindrical space colony 8 km in diameter and 30 km long has been proposed as living quarters for future space explorers.Such a habitat would have cities,land and lakes on the inside surface and air and clouds in the center.All this would be held in place by the rotation of the cylinder about the long axis.How fast would such a cylinder have to rotate to produce a 1-g gravitational field at the walls of the cylinder?
A) 0.05 rad/s
B) 0.10 rad/s
C) 0.15 rad/s
D) 0.20 rad/s

Accepted Answer

To produce a 1-g gravitational field at the walls of the cylinder, the centripetal acceleration must equal the gravitational acceleration (9.8 m/s²). Using the formula for centripetal acceleration a = ω²r, where ω is the angular velocity and r is the radius (4 km or 4000 m), we solve for ω: ω = sqrt(a/r) = sqrt(9.8/4000) ≈ 0.05 rad/s.

Question 19

Somewhere between the Earth and the Moon is a point where the gravitational attraction of the Earth is canceled by the gravitational pull of the Moon.The mass of the Moon is 1/81 that of the Earth.How far from the center of the Earth is this point?
A) 8/9 the way to the Moon
B) 9/10 the way to the Moon
C) 3/4 the way to the Moon
D) 80/81 the way to the Moon

Accepted Answer

The point where the gravitational forces of the Earth and the Moon cancel each other out is known as the Lagrange point L1. The formula to find the distance $r$ from the Earth to L1 in terms of $R$, the distance from the Earth to the Moon, and the masses of the Earth ($M_e$) and the Moon ($M_m$), is $r = R \left(\frac{M_m}{M_m + M_e}\right)^{1/3}$. Given that $M_m = \frac{1}{81}M_e$, substituting this into the formula gives $r = R \left(\frac{1}{1 + 81}\right)^{1/3} = R \left(\frac{1}{82}\right)^{1/3}$. This fraction is approximately 1/9 of the distance from the Earth to the Moon, meaning the point is 1 - 1/9 = 8/9 or approximately 9/10 of the way to the Moon from the Earth, not exactly but closer to 9/10 than to any other options provided.

Question 20

An object of mass 0.50 kg is transported to the surface of Planet X where the object's weight is measured to be 20 N.The radius of the planet is 4.0 × 10⁶ m.What is the mass of Planet X? (G = 6.67 × 10−¹¹ N ⋅ m²/kg²) A) 13 × 10¹⁹ kg B) 17 × 10²² kg C) 9.6 × 10²⁴ kg D) 21 × 10²⁵ kg

Accepted Answer

Using the formula for weight, we have:

W = mg

where W is the weight, m is the mass, and g is the acceleration due to gravity.

On Planet X, we have:

W = 20 N
m = 0.50 kg

We need to find the mass of Planet X, which we can do using the formula for gravitational force:

F = G (m1m2)/r^2

where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of two objects, and r is the distance between them.

On the surface of Planet X, m1 is the mass of the planet, and m2 is the mass of the object (0.50 kg):

F = 20 N
G = 6.67 × 10−11 N⋅m^2/kg^2
m1 = ?
m2 = 0.50 kg
r = 4.0 × 10^16 m

Solving for m1, we get:

m1 = Fr^2/Gm2

m1 = (20 N)(4.0 × 10^16 m)^2/(6.67 × 10−11 N⋅m^2/kg^2)(0.50 kg)

m1 = 9.6 × 10^24 kg

Therefore, the best choice is C, 9.6 × 10^24 kg.

Quiz 7: Rotational Motion and the Law of Gravity

You are facing east looking at the clock on the wall.You see the second hand moving clockwise (of course) .In what direction is the angular velocity vector for this second hand?

What angular speed (in revolutions/second) is needed for a centrifuge to produce an acceleration of 1 000 g at a radius arm of 15.0 cm?

An object of mass 0.50 kg is transported to the surface of Planet X where the object's weight is measured to be 20 N.The radius of the planet is 4.0 × 10⁶ m.What free fall acceleration will the 0.50-kg object experience when at the surface of Planet X?

A roller coaster,loaded with passengers,has a mass of 2 000 kg;the radius of curvature of the track at the bottom point of the dip is 24 m.If the vehicle has a speed of 18 m/s at this point,what force is exerted on the vehicle by the track? (g = 9.8 m/s²)

Consider a child who is swinging.As she reaches the lowest point in her swing:

If a planet has 3 times the radius of the Earth,but has the same density as the Earth,what is the gravitational acceleration at the surface of the planet? (g = 9.8 m/s²)

Consider a point on a bicycle tire that is momentarily in contact with the ground as the bicycle rolls across the ground with constant speed.The direction for the acceleration for this point at that moment is:

The Earth is 93 million miles (mi) from the Sun and its period of revolution is 1 year = 3.15 × 10⁷ s.What is the acceleration of the Earth in its orbit about the Sun?

If a planet has a radius 20% greater than that of the Earth but has the same mass as the Earth,what is the acceleration due to gravity at its surface?

An object when orbiting the Earth at a height of three Earth radii from the center of the Earth has a weight of 1.00 N.What is the object's mass? (g at the surface of the Earth is 9.8 m/s²)

The acceleration due to gravity at the surface of Planet X is 10 m/s².What is the acceleration due to gravity at an altitude of 3 000 km above the surface of this planet?

An airplane in a wide sweeping "outside" loop can create zero gees inside the aircraft cabin.What must be the radius of curvature of the flight path for an aircraft moving at 150 m/s to create a condition of "weightlessness" inside the aircraft?

A wheel is rotated about a horizontal axle at a constant angular speed.Next it is rotated in the opposite direction with the same angular speed.The acceleration at a point on the top of the wheel in the second case as compared to the acceleration in the first case:

An Earth satellite is orbiting at a distance from the Earth's surface equal to one Earth radius (4 000 miles) .At this location,the acceleration due to gravity is what factor times the value of g at the Earth's surface?

At what speed will a car round a 52-m-radius curve,banked at a 45° angle,if no friction is required between the road and tires to prevent the car from slipping? (g = 9.8 m/s²)

At what angle (relative to the horizontal) should a curve 52 m in radius be banked if no friction is required to prevent the car from slipping when traveling at 12 m/s? (g = 9.8 m/s²)

Somewhere between the Earth and the Moon is a point where the gravitational attraction of the Earth is canceled by the gravitational pull of the Moon.The mass of the Moon is 1/81 that of the Earth.How far from the center of the Earth is this point?

An object of mass 0.50 kg is transported to the surface of Planet X where the object's weight is measured to be 20 N.The radius of the planet is 4.0 × 10⁶ m.What is the mass of Planet X? (G = 6.67 × 10−¹¹ N ⋅ m²/kg²)

Quiz 7: Rotational Motion and the Law of Gravity

You are facing east looking at the clock on the wall.You see the second hand moving clockwise (of course) .In what direction is the angular velocity vector for this second hand?

What angular speed (in revolutions/second) is needed for a centrifuge to produce an acceleration of 1 000 g at a radius arm of 15.0 cm?

An object of mass 0.50 kg is transported to the surface of Planet X where the object's weight is measured to be 20 N.The radius of the planet is 4.0 × 106 m.What free fall acceleration will the 0.50-kg object experience when at the surface of Planet X?

A roller coaster,loaded with passengers,has a mass of 2 000 kg;the radius of curvature of the track at the bottom point of the dip is 24 m.If the vehicle has a speed of 18 m/s at this point,what force is exerted on the vehicle by the track? (g = 9.8 m/s2)

Consider a child who is swinging.As she reaches the lowest point in her swing:

If a planet has 3 times the radius of the Earth,but has the same density as the Earth,what is the gravitational acceleration at the surface of the planet? (g = 9.8 m/s2)