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nikdorinn [45]
3 years ago
13

In the real world, orbiting objects such as satellites and stations do not need to be constantly accelerating. What is different

about the real world satellites and the theoretical satellites we are discussing here?
Physics
1 answer:
JulijaS [17]3 years ago
8 0
What do you mean by theoretical satellites? When you mean real world satellites are you talking about the ones in space because I'm not sure what else you talking about.. I wanna help but could you explain plz??
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A walkman uses four standard 1.5 V batteries. How much resistance is in the circuit if it uses a current of 0.02A? *
hodyreva [135]

Answer:

75ohms

Explanation:

V= IR

V = 1.5volts

I = 0.02A

1.5 = 0.02×R

Making R the subject

R = 1.5/0.02

R = 75ohms

The resistance in the circuit will be 75ohms

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When you have been swimming swimming and you come out of the pool, you may feel cold. Use your understanding of endothermic proc
Leona [35]

Explanation:

Edothermic process is the absorbtion of heat.

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4 0
3 years ago
For the PE formula, why is the height required for calculations? Why do we need to know the height in order to determine PE? *
Fudgin [204]

Answer:

Answer in Explanation

Explanation:

Whenever we talk about the gravitational potential energy, it means the energy stored in a body due to its position in the gravitational field. Now, we know that in the gravitational field the work is only done when the body moves vertically. If the body moves horizontally on the same surface in the Earth's Gravitational Field, then the work done on the body is considered to be zero. Hence, the work done or the energy stored in the object while in the gravitational field is only possible if it moves vertically. This vertical distance is referred to as height. <u>This is the main reason why we require height in the P.E formula and calculations.</u>

The derivation of this formula is as follows:

Work = Force * Displacement

For gravitational potential energy:

Work = P.E

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Displacement = Vertical Displacement = Height = h

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5 0
3 years ago
An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 2060 × 103 seconds (about
Fittoniya [83]

Answer:

Acceleration, a=2.22\times 10^{-3}\ m/s^2

Explanation:

It is given that,

Time period of revolution of the moon, T=2060\times 10^3\ s

If the distance from the center of the moon to the surface of the planet is, h=235\times 10^6\ m

The radius of the planet, r=3.9\times 10^6\ m

Let a is the moon's radial acceleration. Mathematically, it is given by :

a=R\times \omega^2, R is the radius of orbit

Since, \omega=\dfrac{2\pi}{T}

The radius of orbit is,

R=r+h

R=3.9\times 10^6\ m+235\times 10^6\ m=238900000\ m

So, a=\dfrac{4\pi^2 R}{T^2}

a=\dfrac{4\pi^2 \times 238900000}{(2060\times 10^3)^2}

a=2.22\times 10^{-3}\ m/s^2

Hence, this is the required solution for the radial acceleration of the moon.

5 0
3 years ago
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