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3 years ago

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A worker pushes a box with a horizontal force of 40.0 N over a level distance of 4.0 m. If a frictional force of 27 N acts on th

e box in a direction opposite to that of the worker, what net work is done on the box

1 answer:

Lorico [155]3 years ago

7 0

Work = (force) x (distance)

The worker does (40N) x (4m) = 160 joules of work.

Friction eats up (27N) x (4m) = 108 joules of that energy,
generating 108 joules of heat.

The remaining (160J - 108J) = 52 joules of energy moves the box.

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If a car crashes into another car like this, the wreck should go nowhere. Besides this being an unrealistic question, the physics of it would look like this:

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Momentum before the collision:
p = m * v = 50000kg * 24m/s + 55000kg * 0m/s = 50000kg * 24m/s
Momentum after the collision:
p = m * v = (50000kg + 55000kg) * v

Setting both momenta equal:
50000kg * 24m/s = (50000kg + 55000kg) * v

Solving for the velocity v:
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A 70.9-kg boy and a 43.2-kg girl, both wearing skates face each other at rest on a skating rink. The boy pushes the girl, sendin

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Answer:

Explanation:

Given

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Suppose speed of boy after push is $v_b$

initially momentum of system is zero so final momentum is also zero because momentum is conserved

$P_i=P=f$

$0=m_b\cdot v_b+m_g\cdot v_g$

$v_b=-\frac{m_g}{m_b}\times v_g$

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i.e. velocity of boy is 2.82 m/s towards west

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Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

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$m_s = 1.989*10^{30}$

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$r = 149.6*10^9m$

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Therefore:

$m = 8.2*1.989*10^{30}$

$d = 6.2*149.6*10^6$

Substituting in Kepler's third law:

$T^2 = \frac{4\pi^2 d^3}{2}$

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Afina-wow [57]

Answer:

Explanation:

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a = 2.06 m /s²

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m = 765 / 9.8

= 78.06 kg

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765 - F = 78.06 x 2.06

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