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2 years ago

An object is moved 29 meters with a force of 289 N. what is the work done

Physics

1 answer:

Shtirlitz [24]2 years ago

8 0

Answer:

8381

Explanation:

Workdone=Force x distance

W= 289x 29

W=8381J

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A soccer ball collides with another soccer ball at rest. the total momentum of the balls

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Two 125 kg bumper cars are moving toward each other in opposite directions. Car X is moving at 10 m/s and Car Z at −12 m/s when

nignag [31]

<h2>Given that,</h2>

Mass of two bumper cars, m₁ = m₂ = 125 kg

Initial speed of car X is, u₁ = 10 m/s

Initial speed of car Z is, u₂ = -12 m/s

Final speed of car Z, v₂ = 10 m/s

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$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

v₁ is the final speed of car X.

$m_1u_1+m_2u_2-m_1v_1=m_2v_2\\\\m_2v_2=m_1u_1+m_2u_2-m_2v_2\\\\m_1v_1=125\times 10+125\times (-12)-125\times 10\\\\v_1=\dfrac{-1500}{125}\\\\v_1=-12\ m/s$

So, car X will move with a velocity of -12 m/s.

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2 years ago

Which two surfaces would have the highest coefficient of friction?

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<span>1. sandpaper against wood.
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2 years ago

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A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$

Thus, the final velocity of thrower is $3.88~m/s$ and that for the catcher is $0.029~m/s$ .

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