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hodyreva [135]
1 year ago
14

1) A satellite in orbit around Earth is in uniform circular motion. What is the angle between the velocity vector and accelerati

on vector at any given point on the orbit?(1 point)
0º
0º

45º
45º

180º
180º

90º

2) A hammer thrower spins a hammer in a circle. What happens when the centripetal force drops to 0?(1 point)

The centripetal force cannot drop to 0.
The centripetal force cannot drop to 0.

The hammer continues in the circular motion.
The hammer continues in the circular motion.

The hammer drops straight to the ground.
The hammer drops straight to the ground.

The hammer flies off at a tangent to its circular path.
3) A bike racer follows a circular indoor track, racing at constant speed. In what direction is their acceleration, and why?(1 point)

In the direction of motion, accelerating their speed.
In the direction of motion, accelerating their speed.

Counter to the direction of motion, keeping her speed constant.
Counter to the direction of motion, keeping her speed constant.

Inward to the center of the track, the direction of the force needed to keep her on this path.
Inward to the center of the track, the direction of the force needed to keep her on this path.

Outward, the direction she would go if not constrained by the track.
4)A diagram shows the velocity vector for an object in uniform circular motion. Why is the vector tangential to the circle?(1 point)

The vector approximates the curved path at that point.
The vector approximates the curved path at that point.

The vector indicates the direction of the force on the object.
The vector indicates the direction of the force on the object.

The vector shows the path the object would follow if the net force acting on it stopped.
The vector shows the path the object would follow if the net force acting on it stopped.

The vector shows how the object pulls against the centripetal force.
5) A satellite is in uniform circular motion around the Moon. What happens to the direction of the acceleration vector during one full orbit?(1 point)

The acceleration vector always points in the direction of motion.
The acceleration vector always points in the direction of motion.

The acceleration vector always points outward.
The acceleration vector always points outward.

The acceleration vector oscillates back and forth.
The acceleration vector oscillates back and forth.

The acceleration vector always points inward.
Physics
1 answer:
artcher [175]1 year ago
6 0

Answer:90

Explanation: bc i looked it up

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Explanation:

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7 0
3 years ago
A box of mass 50 kg is pushed hard enough to set it in motion across a flat surface. Then a 99-N pushing force is needed to keep
vitfil [10]

Answer:

0.20

Explanation:

The box is moving at constant velocity, which means that its acceleration is zero; so, the net force acting on the box is zero as well.

There are two forces acting in the horizontal direction:

- The pushing force: F = 99 N, forward

- The frictional force: F_f=\mu mg, backward, with

\mu coefficient of kinetic friction

m = 50 kg mass of the box

g = 9.8 m/s^2 gravitational acceleration

The net force must be zero, so we have

F-F_f = 0

which we can solve to find the coefficient of kinetic friction:

F-\mu mg=0\\\mu = \frac{F}{mg}=\frac{99 N}{(50 kg)(9.8 m/s^2)}=0.20

7 0
4 years ago
In a catapult, energy is transferred into useful kinetic energy stores. What store is the energy in when it is the catapult?
yawa3891 [41]

Answer:

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6 0
3 years ago
-A city and an international airport are 46 km away from each other. A maglev train takes 11
Wittaler [7]

Answer:

Explanation:

Given the following data;

Distance = 46 km

Time = 11 minutes

To find the average speed;

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7 0
3 years ago
A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.400 rev
saw5 [17]

Answer:

The final angular velocity is rev/s is 0.293 rev/s.

Explanation:

Given;

mass of the merry-go-round, m₁ = 120 kg

radius of the merry-go-round, r = 1.8 m

initial angular velocity, ω = 0.4 rev/s

mass of the child, m₂ = 22 kg

Apply the principle of conservation angular momentum to determine the final angular velocity;

I_i= I_f\\\\\frac{1}{2} m_1r^2 \omega _i = \frac{1}{2} m_1r^2 \omega _f + m_2r^2 \omega _f\\\\ \frac{1}{2} m_1r^2 \omega _i =( \frac{1}{2} m_1r^2  + m_2r^2 )\omega _f\\\\\omega _f = \frac{ \frac{1}{2} m_1r^2 \omega _i}{\frac{1}{2} m_1r^2  + m_2r^2} \\\\\omega _f = \frac{ \frac{1}{2} m_1 \omega _i}{\frac{1}{2} m_1  + m_2}\\\\\omega _f = \frac{0.5 \ \times \ 120\ kg \ \times \ 0.4\ rev/s}{0.5 \ \times 120\ kg \ \ + \ \ 22 \ kg} \\\\\omega _f = 0.293 \ rev/s\\

Therefore, the final angular velocity is rev/s is 0.293 rev/s.

3 0
3 years ago
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