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Sedbober [7]
3 years ago
12

When is an object moving in uniform circular motion?

Physics
1 answer:
Tems11 [23]3 years ago
4 0

Answer: An object undergoing uniform circular motion is moving

Explanation:

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Which of the following mechanical waves has the most energy?
zhenek [66]
Gamma radiation has the greatest energy<span>. This </span>is<span> because gamma radiation </span>has the highest <span>frequency. </span>Energy<span> a frequency.

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kwrob</span>
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Explain why friction makes it harder to push a box filled with groceries than an empty box
kupik [55]
It is harder to push a box with groceries in it because the mass is weighing the box down. Which makes it harder to push
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Work output of a large machine in a factory is 89,000 joules, and it’s input is 102,000 joules. Work output of a similar machine
mojhsa [17]
(89000/102000)×100
=87.25%

(92000/104000)×100
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What would happen to the wavelength if the frequency was doubled?
ira [324]

Answer:

λ = hv

If frequency is doubled :

λ = h × 2v

λ = 2hv

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7 0
3 years ago
When a fixed amount of ideal gas goes through an isobaric expansion A) its internal (thermal) energy does not change.B) the gas
Bingel [31]
<h2>Answer: its temperature must increase.</h2>

Explanation:

In an isobaric process the pressure remains constant, which means the initial pressure and the final pressure will be the same.

In addition, during this thermodynamic process, the volume of the ideal gas expands or contracts in such a way that the variation of pressure \Delta P is neutralized.

Now, according to the First law of Thermodynamics that establishes the conservation of energy:

\Delta U=\Delta Q-\Delta W   (1)

Where:

\Delta U is the internal energy

\Delta Q is the heat transferred

\Delta W is the work

Now, for an isobaric process:

\Delta W=P\Delta V    (2)

Where:

P is the pressure (<u>always positive</u>)

\Delta V is the volume variation of the gas

<u />

<u>Here we have two possible results:</u>

-If the gas expands (positive \Delta V), the work is positive.

-If the gas compresses (negative \Delta V), the work is negative.

In this case we are talking about the first result (work is positive).

Then, according to the above, equation (1) can be written as follows:

\Delta U=\Delta Q - P\Delta V   (3)

Clearing \Delta Q:

\Delta Q=\Delta U+P \Delta V    (4)

Then, for an ideal gas in an isobaric process, part of the heat (Q) added to the system will be used to do work (positive in this case) and the other part <u>will increase the internal energy</u>, hence <u>the temperature will increase as well.</u>

7 0
3 years ago
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