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Sholpan [36]
4 years ago
8

An object has a weight of 500 on earth what is the mass of this object

Physics
1 answer:
erastova [34]4 years ago
7 0

550! OBVY! lol! ope this helps1

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Consider a satellite in a circular orbit around the earth. Why is it important to give a satellite a horizontal speed when placi
Andrei [34K]

Answer:

In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero,

Explanation:

When a satellite is in orbit the most important force is the docking of gravity with the Earth

             F = m a

where the acceleration is centripetal and F is the force of universal attraction

centripetal acceleration is

             a = v² / r

             

             F = m v² / r

In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero, the force also drops to serious and the satellite steels to Earth.

The speed of the satellite is provides the speed, by local for smaller speeds in satellite, it descends in its orbits and when the speed is amate you have the energy to stop an orb to go to a higher orbit.

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3 years ago
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Lynna [10]

Answer:

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Explanation:

8 0
4 years ago
What are ways to improve the design of this experiment? Check all that apply
Serhud [2]

Answer:

B,D,E

Explanation:

I got you

B. Experiment with a wider range of materials.

D.Use a laboratory galvanometer to make precise measurements.

E. Test the strength of the electromagnet by varying the number of wire coils.

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3 years ago
What activity is mainly a strength training excerise?
Sever21 [200]

Biceps curls & pushups , benchpress

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When the distance between two charges is halved, the electrical force between them?
Llana [10]
If the distance between two charges is halved, the electrical force between them increases by a factor 4.

In fact, the magnitude of the electric force between two charges is given by:
F= k \frac{q_1 q_2}{r^2}
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
r'= \frac{r}{2}
the magnitude of the force changes as follows:
F'=k \frac{q_1 q_2}{r'^2}=k \frac{q_1 q_2}{( \frac{r}{2})^2 }=k \frac{q_1 q_2}{ \frac{r^2}{4} } =4k  \frac{q_1 q_2}{r^2}=4 F
so, the force increases by a factor 4.
3 0
3 years ago
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