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Pachacha [2.7K]
2 years ago
7

Anna is sitting in a moving cart and throws a ball straight up. Theoretically, the ball should land in the cart, but it lands on

the ground a little before the cart. What is the reason for the difference in the ranges of the ball and the cart? There is no external horizontal force acting on the ball or the cart. There is no external vertical force acting on the ball or the cart. There is a difference in the external horizontal forces acting on the ball and the cart. There is a difference in the external vertical forces acting on the ball and the cart.
Physics
2 answers:
anastassius [24]2 years ago
8 0

The guys before me was almost right. But instead of it being vertical it is horizontal.

Temka [501]2 years ago
8 0

Answer:

There is a difference in the external horizontal forces acting on the ball and the cart.

Explanation:

When ball is thrown upwards from a moving cart then in that case the acceleration of the ball in vertical direction is due to gravity so it will raised up and again come back to ground.

While during this vertical motion it will also move horizontally with initial speed of cart at which it was moving at the moment when ball is thrown.

Now if ball is not landed on the cart again then we can say that both cart and ball are not accelerated with same acceleration in horizontal direction due to which they both will cover different distances in horizontal direction for the same time interval

So we can say that correct answer will be

There is a difference in the external horizontal forces acting on the ball and the cart.

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Answer:

1) L = 299.88 kg-m²/s

2) L = 613.2 kg-m²/s

3) L = 499.758 kg-m²/s

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6) v = 1.2535 m/s

7) ω₀ = 1.53 rad/s

Explanation:

Given

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We use the equation

L = I₀*ω₀ = 196 kg-m²*1.53 rad/s = 299.88 kg-m²/s

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We use the equation

L = m*v*Rp = 73 kg*4.2 m/s*2.00 m = 613.2 kg-m²/s

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We use the equation

L = m*v*R = 73 kg*4.2 m/s*1.63 m = 499.758 kg-m²/s

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where

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⇒  I₀*ω₀ = (I₀ + m*R²)*ω₁

Now, we can get ω₁

⇒  ω₁ = I₀*ω₀ / (I₀ + m*R²)

⇒  ω₁ = 196 kg-m²*1.53 rad/s / (196 kg-m² + 73 kg*(1.63 m)²)

⇒  ω₁ = 0.769 rad/s

5) Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?

We have to get the centripetal force as follows

Fc = m*ω²*R  

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What is the linear velocity of the person right as they leave the merry-go-round?

we can use the equation

v = ω₁*R = 0.769 rad/s*1.63 m = 1.2535 m/s

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Answer:

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Output power of the circuit is 3 Watt.

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