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2 years ago

7

What are the forces that act on the ball pushed on the floor?Plss answer

1 answer:

vaieri [72.5K]2 years ago

4 0

Explanation:

The ball, for example, will feel gravity pulling it downward and the ground pushing it upward in the direction it is rolling. (Add this if the ball is rolling on the floor.) Friction is the force that causes the ball to slow down because it acts in the opposite direction that it is moving.

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What would be the escape speed for a craft launched from a space elevator at a height of 54,000 km?

Natasha_Volkova [10]

Answer: 3.63 km/s

Explanation:

The escape velocity equation for a craft launched from the Earth surface is:

$V_{e}=\sqrt{\frac{2GM}{R}}$

Where:

$V_{e}$ is the escape velocity

$G=6.67(10)^{-11} Nm^{2}/kg^{2}$ is the Universal Gravitational constant

$M=5.976(10)^{24}kg$ is the mass of the Earth

$R=6371 km=6371000 m$ is the Earth's radius

However, in this situation the craft would be launched at a height $h=54000 km=54000000 m$ over the Eart's surface with a space elevator. Hence, we have to add this height to the equation:

$V_{e}=\sqrt{\frac{2GM}{R+h}}$

$V_{e}=\sqrt{\frac{2(6.67(10)^{-11} Nm^{2}/kg^{2})(5.976(10)^{24}kg)}{6371000 m+54000000 m}}$

Finally:

$V_{e}=3633.86 m/s \approx 3.63 km/s$

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3 years ago

Describe the forces on a rope that supports a tire swing when a child sits on the tire

noname [10]

Potential energy perhaps

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3 years ago

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You have a mass of 60-kg, and you are facing your friend who has a mass of 100 kg. You skate towards

White raven [17]

Answer: My initial velocity is 5 m/s.

Explanation:

In this case, momentum can be conserved.

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Since both the bodies come to rest after collision,

Final momentum = 0

Let my velocity be v, and mass, m1 = 60 kg

Friend's mass, m2 = 100 kg

Friend's velocity, v2 = 3 m/s

Intial momentum = m1v + m2v2

= 60v + 300

Conserving momentum,

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adell [148]

Answer:

Part a)

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Part b)

$y = 3.25 m$

Explanation:

Part a)

Since the diver is moving under gravity

so here its acceleration due to gravity will be uniform throughout the motion

so here we will have

$v_f^2 - v_i^2 = 2 a y$

here we have

$v_i = 2.22 m/s$

$y = -3 m$

$v_f^2 - (2.22)^2 = 2(-9.81)(-3)$

$v_f = 7.99 m/s$

Part b)

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$0 - 2.22^2 = 2(-9.81)(y - 3)$

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3 years ago

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Nutka1998 [239]

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