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Hoochie [10]
3 years ago
7

What occurs when someone places an ice pack on an injured ankle?

Physics
2 answers:
erica [24]3 years ago
7 0

The answer is:

Heat from the ankle is transferred to the ice pack.

The explanation:

-when the ice pack  a lower temp than the ankle , it is very cool.

- and the injured ankle has a high temp or heat

and we know that Q transfer from the object with high heat to the object with the low temp So,

Heat will transfer from the ankle to the ice pack.

Dimas [21]3 years ago
5 0

Answer: 4

Explanation:

Heat from the ankle is transferred to the ice pack.  

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Yes, young rocks typically occur around volcanoes since they are closer to the crate, thus the magma is beginning to cool down to form rocks or it has recently cooled down.

Explanation:

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3 years ago
Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in
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Answer:

94.13 ft/s

Explanation:

<u>Given:</u>

  • t = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s
  • s = distance to be moved by the rock long the horizontal = 98 yards
  • y = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

<u>Assume:</u>

  • u = magnitude of initial velocity of the rock
  • \theta = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\

On dividing equation (1) by (2), we have

\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ

Now, putting this value in equation (2), we have

u\cos 51.34^\circ\times  5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

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