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ZanzabumX [31]
3 years ago
5

A planet in elliptical orbit around a star moves from the point in its orbit furthest from the star (A) to the closest point (P)

. Choose the planet to be your system. The work done by the force of gravity during this movement is:
Physics
1 answer:
Virty [35]3 years ago
6 0

Answer:

Zero work done,since the body isn't acting against  or by gravity.

Explanation:

Gravitational force is usually  considered as work done against gravity (-ve) and work by gravity ( +ve ) and also When work isn't done by or against gravity work done in this case is zero.

Gravitational force can be define as that force that attracts a body to any other phyical body or system that have mass.

The planet been considered as our system in this case is assumed to have mass, and ought to demonstrate such properties associated with gravitational force in such system. Such properties include the return of every object been thrown up as a result of gravity acting downwards. The orbiting nature of object along an elliptical part when gravitational force isn't acting on the body and it is assumed to be zero.

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Before a rifle is fired, the linear momentum of the bullet-rifle system is zero.
marusya05 [52]

Hi there!

II. Linear momentum of the system is zero.

This is an example of a RECOIL collision. With the Law of Conservation of Momentum, momentum remains constant before and after the collision.

Thus, the total momentum would also be equivalent to zero after the collision.

3 0
2 years ago
What are two most important processes in the cycling of oxygen in and out of the atmosphere?
natulia [17]
Weathering and erosion
7 0
3 years ago
What is transferred through sound waves?
alexandr402 [8]

Answer:

Sound waves transfer energy by causing successive compressions and rarefactions in the particles of the medium without transporting the medium particles themselves. Sound in solids can also manifest as transverse waves, causing crests and troughs in the propagation medium.

6 0
3 years ago
A uniform rod of length L is pivoted at L/4 from one end. It is pulled to one side through a very small angle and allowed to osc
ludmilkaskok [199]

Answer:

T= 4.24sec

Explanation:

We are going to use the formula below to calculate.

T=2\pi \sqrt{\frac{L}{g} }

Where T is period

           L is length of rod

       g is acceleration due to gravity =     9.8m/s^{2}

From the problem, the rod is pivoted at 1/4L which means that three quarter of the rod was used for the oscillation. lets call this L_{O}

L_{O} = 3/4 * 5.95m

        = 4.4625m

thus   T=2\pi \sqrt{\frac{L_{O} }{g} }

          T=2\pi \sqrt{\frac{4.4625 }{9.8} }

          T= 4.24sec

8 0
3 years ago
The time constant of an RC circuit is 2.7 s. How much time t is required for the capacitor (uncharged initially) to gain 0.63 of
gayaneshka [121]

Answer:

2.7s

Explanation:

The solution of time required is shown below:-

In the RC circuit condenser charge 63 percent of the full charge from initial time to constant time

Now, the

63% that is equal to 0.63 which is full equilibrium charge

Therefore, the time required to maintain will be Equal to time (t) constant that is 2.7s

So, the correct answer is 2.7s

8 0
3 years ago
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