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3 years ago

If two objects at different temperatures are in contact with each other, what happens to their temperatures?

1 answer:

6 0

In that case, heat energy flows from the warmer object to the cooler one.
As heat flows from one to the other, the temperature of the warmer object
falls, and the temperature of the cooler object rises. When the temperatures
are equal, the flow of heat energy from one to the other stops.

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If vx = 0.40 meters/second and vy = 3.00 meters/second, what is the value of θ?

cestrela7 [59]

<span>θ is the angle whose tangent is 3/0.4 = arctan (7.5) = 82.4°

(That's the angle above the horizontal. If that doesn't match
the </span><span>θ in your diagram, too bad ! For some strange reason,
I wasn't able to see your diagram.)
</span>

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3 years ago

A rifle fires a bullet into a giant slab of butter on a frictionless surface. The bullet penetrates the butter, but while passin

Vera_Pavlovna [14]

Answer:

V= 3.48 m/s

Explanation:

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3 years ago

As the distance between two objects decreases, what happens to the force of gravity between the two objects

KonstantinChe [14]

The force of gravity between them increases.

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4 years ago

Starting from rest, a small block of mass m slides frictionlessly down a circular wedge of mass M and radius R which is placed o

guapka [62]

Answer:

Part a)

$V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

$v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

$y_{cm} = \frac{mR}{m + M}$

Explanation:

PART A)

As we know that there is no external force on the system of two masses in horizontal direction

So here the two masses will have its momentum conserved in horizontal direction

So we have

$mv + MV = 0$

Also we know that here no friction force on the system so total energy will always remains conserved

So we have

$\frac{1}{2}mv^2 + \frac{1}{2}MV^2 = mgR$

now we have

$\frac{1}{2}m(\frac{MV}{m})^2 + \frac{1}{2}MV^2 = mgR$

$\frac{1}{2}MV^2(\frac{M}{m} + 1) = mgR$

so we have

$V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

and another block has speed

$v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

$y_{cm} = \frac{mR}{m + M}$

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3 years ago

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