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sweet [91]
3 years ago
5

Physics help please ;-; . An astronaut has a mass of 82.0 kg. What is the astronaut's stationary weight at a position 4230 kilom

eters above Earth's surface? Note: Earth's radius is 6380 km and the Earth's mass is 5.972 x 1024 kg.
Two rocks, each of mass 72 kg, are positioned 95 m away from each other in deep space. What is the magnitude of the gravitational attraction between them?

if at the least explain how id work it out.​
Physics
1 answer:
sweet [91]3 years ago
3 0

Answer:

F = G*\frac{m_{1} * m_{2} }{r^2}

Explanation:

Just find the force of gravity.

1) m1 is the mass of earth, m2 is mass of astronaut. r is the earth's radius + astronaut's position in meters. G is the universal constant of gravity.

2) Same as above. m1 is 72 kg, m2 is 72 kg, r is 95 m and G is gravitational constant.

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An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.7 s. A passenger in the elevator i
avanturin [10]

Answer: Tension = 53.6N

Explanation:

Given that

Height h = 1 m

Time t = 1.7 s.

Mass m = 5.1 kg 

From the equation of the motion we can get the acceleration of the elevator:

h = X0+ V0t + at2/2;

Th elevator starts from rest with a constant upward acceleration. Initial velocity Vo = 0, also Xo = 0; thus

a = 2h/t2 = 2 × 1/1.7^2

a = 0.69 m/s2.

Then we can find the tension in the cord by using the formula

T = mg + ma

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3 years ago
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Answer:

(a) 1.093 rad/s^2

(b) 4.375 rad/s

(c) 8.744 rad/s

(d)  67.845 rad

Explanation:

initial angular velocity, ωo = 0

time, t = 8s

angular displacement, θ = 35 rad

(a) Let α be the angular acceleration.

Use second equation of motion for rotational motion

\theta =\omega _{0}t+\frac{1}{2}\alpha t^{2}

By substituting the values

35 = 0 + 0.5 x α x 8 x 8

α = 1.093 rad/s^2

(b)  The average angular velocity is defined as the ratio of total angular displacement to the total time taken .

Average angular velocity = 35 / 8 = 4.375 rad/s

(c) Let ω be the instantaneous angular velocity at t = 8 s

Use first equation of motion for rotational motion

ω = ωo + αt

ω = 0 + 1.093 x 8 = 8.744 rad/s

(d) Let in next 5 seconds the angular displacement is θ.

\theta =\omega _{0}t+\frac{1}{2}\alpha t^{2}

By substituting the values

θ = 8.744 x 5 + 0.5 x 1.093 x 5 x 5

θ = 67.845 rad

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