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topjm [15]
2 years ago
5

Suppose you drop a 10-pound weight and a 5-pound weight on the Moon, both from the same height at the same time. What will happe

n? The 10-pound weight will hit the ground before the 5-pound weight. The 5-pound weight will hit the ground before the 10-pound weight. Both weights will float freely because everything is weightless on the Moon. Both will hit the ground at the same time.
Physics
1 answer:
Murljashka [212]2 years ago
3 0

Answer:

Both will hit the ground at the same time.

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Which best compares a brush and a commutator in an electric motor?
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3 years ago
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A cyclist turns a corner with a radius of 50m at a speed of 10m/s. What is the cyclist's acceleration?
miv72 [106K]

2m/s²

Explanation:

Given parameters:

Radius = 50m

Speed = 10m/s

Unknown:

Acceleration of the cyclist

Solution:

The acceleration of the cyclist is directed inside of the corner because his motion is inward. This is a form of centripetal acceleration;

Centripetal acceleration is given by;

     a = \frac{v^{2} }{r}

v is the velocity

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3 years ago
What is the following correct way to write 2,330,000 In a scientific notation
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Answer:

2.33 × 10^6

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8 0
3 years ago
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In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviationMVC. Suppose a 1
lara31 [8.8K]

Answer:

Part a)

f = \frac{8}{9}

Part b)

f = \frac{120}{169}

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

Explanation:

Part a)

Let say the collision between Moose and the car is elastic collision

So here we can use momentum conservation

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1000 v_o = 1000 v_{1f} + 500 v_{2f}

also by elastic collision condition we know that

v_{2f} - v_{1f} = v_o

now we have

2v_o = 2v_{1f} + v_o + v_{1f}

now we have

v_{1f} = \frac{v_o}{3}

Now loss in kinetic energy of the car is given as

\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)

\Delta K = \frac{1}{2}m(v_o^2 - \frac{v_o^2}{9})

so fractional loss in energy is given as

f = \frac{\Delta K}{K}

f = \frac{\frac{4}{9}mv_o^2}{\frac{1}{2}mv_o^2}

f = \frac{8}{9}

Part b)

Let say the collision between Camel and the car is elastic collision

So here we can use momentum conservation

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1000 v_o = 1000 v_{1f} + 300 v_{2f}

also by elastic collision condition we know that

v_{2f} - v_{1f} = v_o

now we have

10v_o = 10v_{1f} + 3(v_o + v_{1f})

now we have

v_{1f} = \frac{7v_o}{13}

Now loss in kinetic energy of the car is given as

\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)

\Delta K = \frac{1}{2}m(v_o^2 - \frac{49v_o^2}{169})

so fractional loss in energy is given as

f = \frac{\Delta K}{K}

f = \frac{\frac{60}{169}mv_o^2}{\frac{1}{2}mv_o^2}

f = \frac{120}{169}

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

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3 years ago
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