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Nikitich [7]
3 years ago
10

For a satellite already in perfect orbit around the Earth, what would happen if: - the satellite's speed is reduced? - the satel

lite's mass is reduced?
Physics
1 answer:
Maksim231197 [3]3 years ago
3 0

Answer:

Answer

Explanation:

If the satellite is in perfect orbit around the earth then gravitational pull of the earth on the satellite and moving forward satellite is in perfect balance i.e gravitational pull does not effect the orbit as the velocity of the satellite is good enough to counter the gravitational pull of the earth and inertia makes the satellite move forward with greater velocity, but when its velocity is suddenly reduced, the gravitational pull will counter on the orbiting satellite and it will crash onto the earth.

This is same as in well of death when motorist moves in circular motion inside the well with centripetal and centrifugal forces balanced, suddenly crashes on the ground when velocity of the motorist is not enough to balance forces.

the satellite's mass has no affect on its orbit around the earth rather the mass of the earth is only affecting factor .

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Nesterboy [21]

They're never the same ray.

4 0
3 years ago
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A long hollow cylindrical conductor (inner radius = 2.0 mm, outer radius = 4.0 mm) carries a current of 12 a distributed uniform
11Alexandr11 [23.1K]

Here we can use ampere'a law to find the magnetic field

\int B.dl = \mu_o i_{en}

B*2 \pi r = \mu_o (i_1 + \frac{i_2*\pi(3^2 - 2^2)}{\pi(4^2 - 2^2}

B*2 \pi*0.003 = 4\pi * 10^{-7} (12 + 5)

B = 4\pi * 10^{-7} (12 + 5)

B = 1.13 * 10^{-3} T

3 0
3 years ago
A car is moving at 25.5 m/s when it accelerates at 1.94 m/s^2 for 2.3 s. What is the car's final speed? (Keep in mind direction
Stolb23 [73]

Answer:

29.96m/s

Explanation:

Given parameters:

Initial speed  = 25.5m/s

Acceleration  = 1.94m/s²

Time  = 2.3s

Unknown:

Final speed of the car  = ?

Solution:

To solve this problem, we are going to apply the right motion equation:

    v = u  + at

v is the final speed

u is the initial speed

a is the acceleration

t is the time taken

 Now insert the parameters and solve;

      v  = 25.5 + (1.94 x 2.3)  = 29.96m/s

3 0
3 years ago
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White raven [17]

Answer:

no the answers r b, d, r, and f

Explanation:

5 0
3 years ago
Read 2 more answers
Two train whistles, and , each have a frequency of 392 Hz. is stationary and is moving toward the right (away from ) at a speed
trasher [3.6K]

Answer:

Part a)

f = 371.1 Hz

Part b)

f = 417.7 Hz

Part c)

beat frequency = 46.6 Hz

Explanation:

Part a)

Due to doppler's Effect the frequency of the sound heard by the train which is moving away from the observer is given as

f_1 = f_0\frac{v + v_o}{v + v_s}

f_1 = 392(\frac{340 + 15}{340 + 35})

f_1 = 371.1 Hz

Part b)

Now from the second train which is approaching the person we can say

f_2 = f_0\frac{v - v_o}{v - v_s}

f_2 = 392(\frac{340 - 15}{340 - 35})

f_2 = 417.7 Hz

Part c)

As we know that beat frequency is the difference in the frequency from two sources

f_b = f_2 - f_1

f_b = 417.7 - 371.1 = 46.6 Hz

8 0
3 years ago
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