Answer:
41.41 m
Explanation:
When force F is applied on an object of mass m for time t and velocity v₁ is created
F X t = mv₁
F = 95 N , t = .53 s, m = 655 kg
95 x .53 = 655 x v₁
v₁ = .0768 m/s
Applying conservation of momentum on man and satellite
m₁ v₁ = m₂v₂
655 x .0768 = 82 xv₂
v₂ = .6134 m/s
their relative velocity
= .6134 + .0768
= .6902 ( they are in opposite direction )
After 60 second distance between them
= 60 x .6902 m
= 41.41 m
The distance between the spacewalking astronaut and the satellite after 1 min is 42.1 m.
The given parameters;
The impulse received by the spacewalking astronaut is equal to the change in momentum of the satellite;
[tex]Ft = m(v-u)\\\\95 \times 0.53 = 655(v-0)\\\\50.35 = 655v\\\\v = \frac{50.35}{655} \\\\v = 0.0768 \ m/s[/tex]
Apply the principle of conservation of linear momentum;
[tex]m_1v_1 = m_2v\\\\82v_1 = 655\times 0.078\\\\82v_1 = 51.09\\\\v _1 = \frac{51.09}{82} \\\\v_1 = 0.623 \ m/s[/tex]
The relative velocity of the spacewalking astronaut and satellite is calculated as;
[tex]v_r = v_1 + v\\\\v_r = 0.623 + 0.078\\\\v_r = 0.701 \ ms/[/tex]
The distance between the spacewalking astronaut and the satellite after 1 min is calculated as;
[tex]d = v_rt\\\\d = 0.701 \times 60\\\\d = 42.1 \ m[/tex]
Thus, the distance between the spacewalking astronaut and the satellite after 1 min is 42.1 m.
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