What is the magnitude of the normal force?

A satellite of mass 5600 kg orbits the Earth and has a period of 6200 s.

Troyanec [42]

Answer:

(a) Radius of orbit will be $=7.32\times10^6m$

(b) Earth gravitational force will be $=4.18\times 10^4N$

(C) Height will be $0.92\times 10^6m$

Explanation:

We have given

Mass of the earth, $M=6\times 10^{24}kg$

Mass of the satellite, m = 5600 kg

Radius of earth, $R=6.4\times 10^6m$

Time period T = 6200 sec

We know that $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{6200}=0.00101rad/sec$

Now

(a) We know that $\omega ^2=\frac{GM}{R^3}$

$R^3=\frac{GM}{\omega ^2}$

$R^3=\frac{6.67\times 10^{-11}\times 6\times 10^{24}}{0.00101 ^2}$

$R^3=3.92\times 10^{20}$

Radius of the orbit $R=7.32\times 10^6m$

(b)

Force $F=\frac{GMm}{R^2}=\frac{6.67\times 10^{-11}\times 6\times 10^{24}\times 5600}{(7.32\times 10^6)^2}=4.18\times 10^4N$

(c)

Altitude $h=radius\ of\ orbit-radius\ of\ earth=7.32\times 10^6-6.4\times 10^6=0.92\times 10^6m$

8 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

Luden [163]

Answer:

The ball will be attracted to the negatively charged plate. It'll touch and pick up some electrons from the plate so that the ball becomes negatively charged. Immediately the ball is repelled by the negative plate and is attracted to the positive plate. The ball gives up electrons to the positive plate so that it is positively charged and suddenly attracts to the negative plate again, flies over to it and picks up enough electrons to be repulsed by negative plate and again to the positive plate and that continues.

8 0

3 years ago

If you walk 5 km north and then 12 km east. Your resultant displacement is____.a. magnitude: 1.1km; direction: 53.1 degrees east

sweet [91]

Answer:

Resultant = 13km

Direction = 67.38° East of North

Explanation:

Given the following :

5km North ; 12km East

Resultant Displacement (r) :

r² = 5² + 12²