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Kryger [21]
3 years ago
6

A satellite with a mass of 2,000 ㎏ is inserted into an orbit that is twice the Earth' s radius. W hat is the force of gravity on

the satellite? The radius of the Earth is 6380 km, and the mass of the Earth is 5.98 X 1024 kg a. 2,000 N h. 4,900 N 19,600 N d 24,500 N e 36.200 N l Space Station which orbits Earth, astronauts experience apparenst
Physics
1 answer:
dolphi86 [110]3 years ago
4 0

Answer:

option (b) 4900 N

Explanation:

m = 2000 kg, R = 6380 km = 6380 x 10^3 m, Me = 5.98 x 10^24 kg, h = R

F = G Me x m / (R + h)^2

F = G Me x m / 2R^2

F = 6.67 x 10^-11 x 5.98 x 10^24 x 2000 / (2 x 6380 x 10^3)^2

F = 4900 N

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A ball on a string makes 25.0 revolutions in 9.37 s, in a circle radius 0.450 m. What is it’s velocity?
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The velocity of the ball is 12.5 m/s

Explanation:

The velocity of the ball is given by the ratio between the distance covered by the ball and the time taken:

v=\frac{d}{t}

First, we calculate the distance covered. We know that the radius of the circle is

r = 0.450 m

And the length of the circumference is

L=2\pi r = 2\pi(0.750)=4.7 m

The ball makes 25.0 revolutions, so a total distance of

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In a time of

t = 9.37 s

So, its velocity is

v=\frac{117.5}{9.37}=12.5 m/s

Learn more about velocity here:

brainly.com/question/5248528

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A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo
Rasek [7]

Answer:

t=12.25\ seconds

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<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

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\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}

Where vo is the magnitude of the initial velocity, \theta is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

\displaystyle t=\frac{v_osin\theta}{g}

There are two times where the value of y is y_o when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making y=y_o

\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}

Removing y_o and dividing by t (t different of zero)

\displaystyle 0=v_osin\theta-\frac{gt}{2}

Then we find the total flight as

\displaystyle t=\frac{2v_osin\theta}{g}

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

\boxed{t=24.5/2=12.25\ seconds}

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The velocity of any object depends upon
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The displacement, time, and direction
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