The speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.
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Speed of the satellite</h3>
v = √GM/r
where;
- M is mass of Earth
- G is universal gravitation constant
- r is distance from center of Earth = Radius of earth + 4930 km
v = √[(6.626 x 10⁻¹¹ x 5.97 x 10²⁴) / ((6371 + 4930) x 10³)]
v = 5,916.36 m/s
Thus, the speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.
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The gravitational potential energy
gpe = mgh
Answer:
96 m
Explanation:
Given,
Initial velocity ( u ) = 4 m/s
Final velocity ( v ) = 20 m/s
Time ( t ) = 8 s
Let Acceleration be " a ".
Formula : -
a = ( v - u ) / t
a = ( 20 - 4 ) / 8
= 16 / 8
a = 2 m/s²
Let displacement be " s ".
Formula : -
s = ut + at² / 2
s = ( 4 ) ( 8 ) + ( 2 ) ( 8² ) / 2
= 32 + ( 2 ) ( 64 ) / 2
= 32 + ( 2 ) ( 32 )
= 32 + 64
s = 96 m
Therefore, it travels 96 m in time 8 s.
A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.
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Amplitude: How dense the medium is in the compression part of the wave, and how empty the rarefied area is.
Frequency: The number of wavelengths that pass a position in 1 second.
loudness: The quality of the sound that is most closely linked to the amplitude of the sound wave.
Period: The amount of time that it takes one wavelength to pass by a position.
Pitch: The quality of the sound that is most closely linked to the frequency of the sound wave.
