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Irina-Kira [14]
3 years ago
5

Can someone please help me with this? I'm so confused

Physics
1 answer:
bagirrra123 [75]3 years ago
3 0

At theheight where it starts, just before it's dropped, the ball has
some potential energy.  The higher that spot is, the more potential
energy the ball has.  After the drop, whenever the ball is lower than
the height from which it was dropped, it has less potential energy, and
the missing potential energy shows up as kinetic energy ... motion.

This is the whole idea of the roller coaster.  A machine drags it up to
the top of the first hill, giving it lots of potential energy.  After that, as
long as it doesn't try to rise higher than the first hill, it never runs out
of energy, and keeps going.

A). and B).
The ball keeps going forward until it rises again to the same height it
was dropped from ... on the other side.  Then it stops and falls back.

C). The ball can never rise higher than the height it was dropped from.
If the hump in the middle is the same height as the drop-height, then
the ball stops right there, and falls back.

D).  Same as B).  As long as the track inside the loop is never higher
than the droop-height, the ball just keeps going forward.

E). Same idea.  Here it looks like the drop-height is the same as the
top of the loop. The ball can't rise higher than it was dropped from,
so it gets as far as the top of the loop and stops there.  From there,
I think it drops straight down from the top of the loop, instead of
following the curve.


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A 26.4 g silver ring (cp = 234 J/kg·°C) is heated to a temperature of 66.2°C and then placed in a calorimeter containing 4.94 ✕
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The final temperature of the mixture = 64.834 °C.

Explanation:

Heat lost by the silver ring = heat gained by the water + heat transferred to the surrounding.

c₁m₁(t₁-t₃) = c₂m₂(t₃-t₂) + Q..............Equation 1

Where c₁ = specific heat capacity of the silver copper, m₁ = mass of the silver copper, t₁ = initial temperature of the silver copper, t₃ = final temperature of the mixture. c₂ = specific heat capacity of water, t₂ = initial temperature of water, m₂ = mass of water, Q = energy transferred to the surrounding.

making t₃ the subject of the equation,

t₃ = [c₁m₁t₁+c₂m₂t₂-Q]/(c₁m₁+c₂m₂)........................ Equation 2

Given: c₁ = 234 J/kg.°C, m₁ = 26.4 g, t₁ = 66.2 °C, c₂ = 4200 J/K.°C, m₂ = 4.92×10⁻² kg, t₂ = 24.0 °C, Q = 0.136 J.

Substituting into equation 2

t₃ = [(234×26.4×66.2)+(4200×0.0492×24)-0.136]/[(234×26.4)+(4200×0.0492)]

t₃ = (408957.12+4959.36-0.136)/(6177.6+206.64)

t₃ = (413916.48-0.136)/6384.24

t₃ = 413916.34/6384.24

Thus the final temperature of the mixture = 64.834 °C.

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Sitting in a chairlift, Rebecca has a gravitational potential energy of 5,997.6
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