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borishaifa [10]

3 years ago

8

A 30-kg child sits at the top of a 3-meter slide. After sliding down, the child is traveling 4 m/s. How much PE does he start wi

th? How much KE does he end with? How much energy is lost to friction?

1 answer:

Jobisdone [24]3 years ago

6 0

At the top:

   Potential Energy = (mass) x (gravity) x (height)

   = (30 kg) x (9.8 m/s²) x (3 meters)

   = 882 joules

At the bottom:

   Kinetic Energy = (1/2) x (mass) x (speed)²

   = (1/2) x (30 kg) x (3 m/s)²

   = (15 kg) x (9 m²/s²)

   = 135 joules .

He had 882 joules of potential energy at the top,
but only 135 joules of kinetic energy at the bottom.

Friction stole (882 - 135) = 747 joules of his energy while he slid down.
The seat of his jeans must be pretty warm.

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On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

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Answer:

$v_a=0.8176 m/s$

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Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is

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Since b is initially at rest

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After the collision and being $v'_a$ and $v'_b$ the respective velocities, the total momentum is

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Both momentums are equal, thus

$m_av_a=m_av'_a+m_bv'_b$

Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

$v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}$

$v_a=0.8176 m/s$

The initial kinetic energy can be found as (provided puck b is at rest)

$K_1=\frac{1}{2}m_av_a^2$

$K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J$

The final kinetic energy is

$K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2$

$K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J$

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A baseball player hits a homerun, and the ball lands in the left field seats, which is 103m away from the point at which the bal

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$y=v_{y,i}t-\dfrac12gt^2$

so that when it lands in the seats at <em>t</em> ≈ 6.38 s, it has a height of

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An ideally efficient heat pump delivers 1000 J of heat to room air at 300 K. If it extracted heat from 260 K outdoor air, how mu

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Answer:

Wnet, in, = 133.33J

Explanation:

Given that

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COP(HP, rev) = 1/(1-260/300)

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Extra....

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