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cluponka [151]
3 years ago
6

A 56-kg skater initially at rest throws a 5.0-kg medicine ball horizontally to the left. Suppose the ball is accelerated through

a distance of 1.0 m before leaving the skater's hand at a speed of 7.0 m/s. Assume the skater and the ball to be point-like and the surface to be frictionless and ignore air resistance. Use a vertical y-axis with the positive direction pointing up and a horizontal x-axis with the positive direction pointing to the right. What will happen to the skater and the ball after the ball is thrown
Physics
1 answer:
aniked [119]3 years ago
4 0

Answer:

a_2=24.5\ \text{m/s}^2 towards right

a_1=2.1875\ \text{m/s}^2 towards left

Explanation:

m_1 = Mass of skater = 56 kg

m_2 = Mass of ball = 5 kg

v = Final velocity of ball = 7 m/s

u = Initial velocity of ball = 0

s = Distance the ball moved in the hand of the skater = 1 m

Moving left is considered and moving right is considered positive.

From kinematic equations of motion we have

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{7^2-0^2}{2\times 1}\\\Rightarrow a=24.5\ \text{m/s}^2

So, the ball will move towards right with a magnitude of acceleration a_2=24.5\ \text{m/s}^2.

The force on the ball will be

F_2=m_2a_2\\\Rightarrow F_2=5\times 24.5\\\Rightarrow F_2=122.5\ \text{N}

The force on the ball is 122.5\ \text{N}

The reaction force on the skater will be equal to the force on the ball but will have opposite direction.

-F_1=F_2\\\Rightarrow -m_1a_1=F_2\\\Rightarrow a_1=-\dfrac{122.5}{56}\\\Rightarrow a_1=-2.1875\ \text{m/s}^2

So, the skater will move towards left with a magnitude of acceleration 2.1875\ \text{m/s}^2

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Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3  m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

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Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

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