Answer:
a) f ’’ = f₀ , b) Δf = 2 f₀
Explanation:
a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.
Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source
f ’= fo
This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.
f ’’ = f’
where c represents the sound velocity in stationary blood
therefore the received frequency is
f ’’ = f₀
let's simplify the expression
f ’’ = f₀ \frac{c+v}{c-v}
f ’’ = f₀
b) At the low speed limit v <c, we can expand the quantity
(1 -x)ⁿ = 1 - x + n (n-1) x² + ...
f ’’ = fo
f ’’ = fo
leave the linear term
f ’’ = f₀ + f₀ 2
the sound difference
f ’’ -f₀ = 2f₀ v/c
Δf = 2 f₀
It changes because force is somewhat like pressure. Force is continuously against the object so as a result, speed changes .
(a) The speed of a satellite on a low lying circular orbit around this planet is 7,338.93 m/s.
(b) The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.
(c) The radius of the synchronous orbit of a satellite is 69,801 km .
<h3>Speed of the satellite</h3>
v = √GM/r
where;
- M is mass of the planet
- r is radius of the planet
v = √[(6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]
v = 7,338.93 m/s
<h3>Escape velocity of the satellite</h3>
v = √2GM/r
v = √[( 2 x 6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]
v = 10,378.82 m/s
<h3>Speed of the satellite at the given period </h3>
v = 2πr/T
r = vT/2π
r = (7,338.93 x 16.6 x 3600 s) / (2π)
r = 69,801 km
Thus, the speed of a satellite on a low lying circular orbit around this planet is 7,338.93 m/s.
The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.
The radius of the synchronous orbit of a satellite is 69,801 km .
Learn more about minimum speed here: brainly.com/question/6504879
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Answer:
<em>The centripetal acceleration will be increased to 1.33 of its initial state. </em>
Explanation:
<h3>Centripetal acceleration</h3>
The Centripetal acceleration of an object is the acceleration of the object along a circular path moving towards the center of the circular path. The centripetal acceleration is represented in the equation bellow
...................................... 1
where is the centripetal acceleration
v is the tangential velocity
and r is the radius.
<h3>How the Change of Radius Affects the Centripetal Acceleration</h3>
Reference to equation 1 the centripetal acceleration ( ) is inversely proportional () to the radius of the circle or path. this means that when the radius increases the centripetal acceleration reduces and when the radius reduces the centripetal acceleration increases. The radius was reduced to 0.75R in the question that will amount to 1.33 increase in the centripetal acceleration. This can be obtained by multiplying the centripetal acceleration by the inverse of 0.75 which is 1.33.
Therefore, when the radius is reduced by 0.75R , the centripetal acceleration of the steel ball will increase by 1.33. since the period is kept constant