Question 1

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

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Question 2

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

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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 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 4

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

This point is called the L1 Lagrange point and is located approximately 9/10 of the way from the center of the Earth to the Moon. The exact position depends on various factors such as the distance between the Earth and Moon, and their relative positions in their orbits. Since the mass of the Moon is much smaller than that of the Earth, the L1 point is closer to the Moon than the Earth.

Question 5

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 x 10^-3 mi/s² C)13.6 x 10^-6 mi/s² D)3.7 x 10^-6 mi/s²

Accepted Answer

To find the acceleration of the Earth in its orbit about the Sun, we can use the following formula:

a = v^2/r

where
a = acceleration
v = velocity
r = radius of orbit

First, we need to find the velocity of the Earth in its orbit. We can use the formula:

v = 2πr/T

where
T = period of revolution

Plugging in the values, we get:

v = 2π(93 x 10^6)/3.15 x 10^7
v = 29.5 mi/s

Now, we can find the acceleration using the first formula:

a = (29.5)^2/(93 x 10^6)
a = 9.32 x 10^-3 mi/s^2
a ≈ 3.7 x 10^-6 mi/s^2

Therefore, the best choice is D, 3.7 x 10^-6 mi/s^2.

Question 6

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 7

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.

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Question 8

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 9

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

The acceleration due to gravity at the surface of a planet is given by $g = \frac{GM}{r^2}$, where $G$ is the gravitational constant, $M$ is the mass of the planet, and $r$ is the radius of the planet. Since the planet in question has the same mass as the Earth, $M$ is the same for both. However, the radius of the planet is 20% greater than that of the Earth, which means $r$ for this planet can be written as $1.2r_E$, where $r_E$ is the radius of the Earth. Substituting these values into the expression for $g$, we get: $$g = \frac{GM}{r^2} = \frac{GM}{(1.2r_E)^2} = \frac{1}{1.44}\left(\frac{GM}{r_E^2}ight) = \frac{1}{1.44}g_E \approx 6.8 	ext{ m/s}^2$$ where $g_E$ is the acceleration due to gravity at the surface of the Earth. Therefore, the best choice is (D).

Question 10

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 11

At what speed will a car round a 52-m-radius curve,banked at a 45 $\degree$ 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

Accepted Answer

The formula to calculate speed in a banked curve with no friction is
v = $\sqrt{rg	an	heta}$
Here, r = 52 m, g = 9.8 m/s$^2$ and $	heta$ = 45$^\circ$ = $\frac{\pi}{4}$ radians.
Substituting the values in the formula, we get
v = $\sqrt{52 \cdot 9.8 \cdot 	an(\frac{\pi}{4})}$ = 22.96 m/s $\approx$ 23 m/s (rounded to nearest whole number)
So, the correct option is C.

Question 12

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 x 10⁶ m.What is the mass of Planet X? (G = 6.67 x 10^-11 N*m²/kg²) A)13 x 10¹⁹kg B)17 x 10²² kg C)9.6 x 10²⁴ kg D)21 x 10²⁵kg

Accepted Answer

The weight of the object on Planet X is given by:
W = G(Mm/r^2)
where M is the mass of the planet, m is the mass of the object, r is the radius of the planet, and G is the gravitational constant.

We can rearrange this equation to solve for M:
M = (Wr^2)/(Gm)

Substituting the given values, we get:
M = (20 N)(4.0 x 10^16 m)^2 / [(6.67 x 10^-11 Nm^2/kg^2)(0.50 kg)]
M = 9.6 x 10^24 kg

Therefore, the mass of Planet X is 9.6 x 10^24 kg. The correct answer is (C).

Question 13

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 14

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

Weight (W) = mg, where m = 0.50 kg and W = 20 N (as given in the problem)

Therefore, g = W/m = 20 N/0.50 kg = 40 m/s^2

Now, we know that the acceleration due to gravity (g) is given by:

g = GM/r^2, where G is the gravitational constant (6.67 × 10^-11 N·m^2/kg^2), M is the mass of the planet, and r is the radius of the planet.

We can rearrange this formula to solve for M:

M = gr^2/G

Substituting the given values, we get:

M = (40 m/s^2) × (4.0 × 10^16 m)^2 / (6.67 × 10^-11 N·m^2/kg^2) = 3.80 × 10^27 kg

Now, we can use the formula for g again to find the acceleration due to gravity at the surface of Planet X, using its mass and radius:

g = GM/r^2 = (6.67 × 10^-11 N·m^2/kg^2) × (3.80 × 10^27 kg) / (4.0 × 10^16 m)^2 ≈ 48 m/s^2

Therefore, the answer is D, with an acceleration due to gravity of 48 m/s^2.

Question 15

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 $\degree$ B)32 $\degree$ C)16 $\degree$ D)10 $\degree$

Accepted Answer

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Question 16

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 17

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 18

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 x 10⁴ N B)4.7 x 10⁴ N C)3.0 x 10⁴ N D)1.0 x 10⁴ N

Accepted Answer

We can use the formula F = (m*v^2)/r to find the force exerted on the vehicle by the track at the bottom of the dip.
F = (2000 kg)*(18 m/s)^2/(24 m)
F = 4.7 x 10^4 N
Therefore, the correct answer is B, 4.7 x 10^4 N.

Question 19

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 20

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.

Quiz 7: Rotational Motion and the Law of Gravity

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)

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?

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?

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)

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?

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?

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 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?

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

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 x 10⁶ m.What is the mass of Planet X? (G = 6.67 x 10^-11 N*m²/kg²)

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?

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)

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:

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

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 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:

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?

Quiz 7: Rotational Motion and the Law of Gravity

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)

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?

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?

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)