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Galina-37 [17]
3 years ago
9

Consider a coin which is tossed straight up into the air. After it is released it moves upward, reaches its highest point and fa

lls back down again. Air resistance can be neglected. What force acts on the coin while it moves upward after it is released
Physics
1 answer:
avanturin [10]3 years ago
3 0

Answer:

GRAVITATIONAL FORCE

Explanation:

We may have noticed that a body thrown upward in air falls back down again after attaining a particular height. The object was able to fall down back due to the effect of gravity acting on it. If there are no force of gravity acting on the body, the body will not fall back but rather disappears into the thin air.

A coin tossed upward in the air which falls back down when released is therefore under the influence of gravity i.e GRAVITATIONAL FORCE while it moves upward after it is released

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Answer:

1.34352 kg

Explanation:

m_w = Mass of water falling = 1 kg

h = Height of fall = 0.1 km

\Delta T = Change in temperature = 0.1

c = Specific heat of water = 4186 J/kg K

g = Acceleration due to gravity = 9.81 m/s²

m_v = Mass of water in the vessel

Here the potential energy will balance the internal energy

m_wgh=m_wc\Delta T+m_vc\Delta T\\\Rightarrow m_v=\dfrac{m_wgh-m_wc\Delta T}{c\Delta T}\\\Rightarrow m_v=\dfrac{m_wgh}{c\Delta T}-m_w\\\Rightarrow m_v=\dfrac{1\times 9.81\times 100}{4186\times 0.1}-1\\\Rightarrow m_v=1.34352\ kg

Mass of the water in the vessel is 1.34352 kg

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3 years ago
PLEASE HELP ME I JUST CAN'T FIGURE THIS ONE OUT
kupik [55]
Indeed the answer is c!!
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Answer:

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