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shusha [124]
3 years ago
14

Estimate the smallest possible period of a satellite in a circular orbit around earth. (mass and radius of earth is 5.98 x 1024

kg and 6.37 x 106 m. the universal gravitational constant g = 6.67 x 10-11 n.m2/kg2 .)
Physics
1 answer:
myrzilka [38]3 years ago
3 0
Given: Mass of earth Me = 5.98 x 10²⁴ Kg

           Radius of earth r = 6.37 x 10⁶ m

            G = 6.67 x 10⁻¹¹ N.m²/Kg²

Required: Smallest possible period T = ?

Formula: F = ma;  F = GMeMsat/r²     Centripetal acceleration ac = V²/r

               but V = 2πr/T

equate T from all equation.

F = ma

GMeMsat/r² = Msat4π²/rT²    

GMe = 4π²r³/T²

T² = 4π²r³/GMe  

T² = 39.48(6.37 x 10⁶ m)³/6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)

T² = 1.02 x 10²² m³/3.99 x 10¹⁴ m³/s²

T² = 25,563,909.77 s²

T = 5,056.08 seconds    or around   1.4 Hour

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natta225 [31]

Answer:

<h2>5850 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 750 × 7.8

We have the final answer as

<h3>5850 N</h3>

Hope this helps you

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How does the frequency of sound relate to its pitch?
madreJ [45]
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3 years ago
A coin is dropped into a wishing well. It takes 1.1 seconds for a splash to be heard. Calculate the depth of the wishing well
Zigmanuir [339]

Answer:

If the wishing well was in a vacuum, then s=ut + 0.5 a t^2 (s=distance, ... wishing well if you drop a coin into it and hear the splash 10 seconds

Explanation:

8 0
2 years ago
Special relativity can be used to study an object in which frame of reference?
Tresset [83]
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3 0
3 years ago
Read 2 more answers
A rock is thrown off a 50.0 m high cliff. How fast must the rock leave the cliff top to land on level ground below, 90 m from th
blagie [28]

Answer:

The rock must leave the cliff at a velocity of 28.2 m/s

Explanation:

The position vector of the rock at a time t can be calculated using the following equation:

r = (x0 + v0x · t, y0 + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the cliff so that x0 and y0 = 0.

When the rock reaches the ground, the position vector will be (see r1 in the figure):

r1 = (90 m, -50 m)

Then, using the equation of the vector position written above:

90 m = x0 + v0x · t

-50 m = y0 + 1/2 · g · t²

Since x0 and y0 = 0:

90 m = v0x · t

-50 m = 1/2 · g · t²

Let´s use the equation of the y-component of the vector r1 to find the time it takes the rock to reach the ground and with that time we can calculate v0x:

-50 m = 1/2 · g · t²

-50 m = -1/2 · 9.81 m/s² · t²

-50 m / -1/2 · 9.81 m/s² = t²

t = 3.19 s

Now, using the equation of the x-component of r1:

90 m = v0x · t

90 m = v0x · 3.19 s

v0x = 90 m / 3.19 s

v0x = 28.2 m/s

8 0
3 years ago
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