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Helen [10]
3 years ago
6

When an object is moved from rest at point A to rest at point B in a gravitational field, the net work done by the field depends

on the mass of the object and
1. the nature of the external force moving the object from A to B.

2. the velocity of the object as it moves between A and B.

3. both the positions of A and B and the path taken between them.

4. the path taken between A and B only.

5. the positions of A and B only.
Physics
1 answer:
9966 [12]3 years ago
4 0

Answer:5. The positions of A and B only.

Explanation: Gravitational force or pull is the pull on all objects by the Earth Magnetic forces,it tends to attract all objects to itself. Gravitational field is a field induced by an object with mass, which will interact with other massive objects by applying a force on it . Gravitational field does not depend on the paths of motion of any object,its velocity, the nature of the forces around the object,the Gravitational force is known to be a conservative force,which tries to conserve its force.

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One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K from a pressure of 10 atm to a pressure of 2 at
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Answer:

Explanation:

Given

1 mole of perfect, monoatomic gas

initial Temperature(T_i)=300 K

P_i=10 atm

P_f=2 atm

Work done in iso-thermal process=P_iV_iln\frac{P_i}{P_f}

P_i=initial pressure

P_f=Final Pressure

W=10\times 2.463\times ln\frac{10}{2}=39.64 J

Since it is a iso-thermal process therefore q=w

Therefore q=39.64 J

(b)if the gas expands by the same amount again isotherm-ally and irreversibly

work done is=P\Delta V

V_1=\frac{RT_1}{P_1}=\frac{1\times 0.0821\times 300}{10}=2.463 L

V_2=\frac{RT_2}{P_2}=\frac{1\times 0.0821\times 300}{2}=12.315 L

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8 0
4 years ago
Ignoring air resistance and the little friction from the plastic tube, the magnet was a freely-falling object in each trial. If
horsena [70]

Answer:

emf will also be 10 times less as compared to when it has fallen 40 \mathrm{m}

Explanation:

We know, from faraday's law-

e m f=-N \frac{\Delta \Phi}{\Delta T}

and \Phi=B . A

So, as the height increases the velocity with which it will cross the ring will also increase. (v=\sqrt{2 g h})

Given

\mathrm{V} 1(\text { Speed at } 40 \mathrm{m})=2 \mathrm{x} \mathrm{V} 2(\text { speed at } 10 \mathrm{m})

\sqrt{2 g h_{2}}=2 \times \sqrt{2 g h_{1}}=28.28 \mathrm{m} / \mathrm{s}

Now, from 40 \mathrm{cm}

V_{3}=\sqrt{2 g h_{3}}=\sqrt{2 \times 10 \times 0.4}=2.82 \mathrm{m} / \mathrm{s}

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