Answer:
The tension in the left side string = 17.21 N
The tension in the right side string = F = 27.3 N
Explanation:
Given that
F= 27.3 N
M= 1.43 kg ,r= 0.0792 m
Moment of inertia of disk ,I = 0.5 m r²
I = 0.5 x 1.43 x 0.0792² = 0.0044 kg.m²
m= 0.7 kg
Lets take linear acceleration of system is a m/s²
Lets take tension in left side string = T
From Newtons law
T- mg = ma
T- 0.7 x 10 = 0.7 a ----------1
(F - T) r = I α
α = Angular acceleration of disk
a = α r
(F - T) r = I α
(F - T) r² = I a
( 27.3 - T) x 0.0792² = 0.0044 a --------2
Form equation 1 and 2
a= 1.42 T - 10 m/s²
a = 1.42 ( 27.3 - T) m/s²
1.42 T - 10 = 38.9 - 1.42 T
T=17.21 N
The tension in the right side string = F = 27.3 N
Acceleration is the rate of change of velocity per unit of time.
(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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Ans: (a) & (c)
The density of a material is its mass per unit volume. This statement is proof enough to validate (c) as an answer. Now 1cm3 is also equal to 1mL, which makes (c)=(a).