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kolezko [41]
2 years ago
9

A 92-kg astronaut and a 2000-kg satellite are at rest relative to a space station. the astronaut pushes on the satellite, giving

it a speed of 0.14 m/s directly away from the station. seven and a half seconds later the astronaut comes into contact with the station. part a what was the initial distance from the station to the astronaut?
Physics
2 answers:
just olya [345]2 years ago
8 0
Momentum is conserved. 

<span>Momentum of the satellite is 1600*0.14=224 kg-m/sec </span>
<span>Resulting momentum of the astronaut will be equal -- hence </span>
<span>86*v=224 or the velocity of the astronaut is 2.60 m/sec </span>

<span>In 7.5 seconds the astronaut would travel 19.53 meters</span>
jekas [21]2 years ago
7 0

Answer:

22.8 s

Explanation:

First we calculate the change in momentum of the satellite.

ΔP = 2000 kg (0.14 m/s 0 m/s) = 280 kg·m/s

The change in momentum of the satellite is equal to the change in momentum of the astronaut as a result of Newton's Third law. Therefore:

280 kg·m/s = 92kg * v

                 v = 3.04 m/s

d = speed * time = 3.04 m/s * 7.5 s = 22.8 m

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find the period of a simple pendulum of 1m length placed on earth and on moon g on moon =1.67m/s² g on earth=10m/s²
Ierofanga [76]

Answer:

T_{m } = 4.86 s

T_{e} = 1.98 s

Explanation:

<u><em>Given:</em></u>

Length = l = 1 m

Acceleration due to gravity of moon = g_{m} = 1.67 m/s²

Acceleration due to gravity of Earth = g_{e} = 10 m/s²

<u><em>Required:</em></u>

Time period = T = ?

<u><em>Formula:</em></u>

T = 2π \sqrt{\frac{l}{g} }

<u><em>Solution:</em></u>

<u>For moon</u>

<em>Putting the givens,</em>

T = 2(3.14) \sqrt{\frac{1}{1.67} }

T = 6.3 \sqrt{0.6}

T = 6.3 × 0.77

T = 4.86 sec

<u>For Earth,</u>

<em>Putting the givens</em>

T = 2π \sqrt{\frac{1}{10} }

T = 2(3.14) \sqrt{0.1}

T = 6.3 × 0.32

T = 1.98 sec

3 0
3 years ago
12. A runner at the start of a race generates 250 W of power as he accelerates to 5 m/s. If the runner has a mass of
Margaret [11]
1/2 m/s2
30 Na




I hope this helps!
4 0
3 years ago
E<br> 3.6 What force is needed to give a mass of<br> 20 kg an acceleration of 5 m/s??
luda_lava [24]

Explanation:

  • Mass(m)= 20kg
  • Acceleration (a)= 5m/s²
  • Force(F)= ?

We know that,

  • F=ma
  • F=20×5
  • F=100N

Hence, the needed force is 100N.

6 0
1 year ago
Read 2 more answers
The 0.15kg baseball has a speed of v=30 m/s just before it is struck by the bat. It then travels along the trajectory shown befo
notka56 [123]

The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N

The mass of the baseball, m = 0.15 kg

The speed at which it moves, v = 30 m/s

Time at which the baseball was in contact with the bat, t = 0.75 ms

t = 0.75/1000 s

t  = 0.00075 s

The impulsive force is given by the formula:

F=\frac{mv}{t}

Substitute m = 0.15 kg, v = 30, and t = 0.00075s into the formula above:

F=\frac{0.15 \times 30}{0.00075} \\\\F=6000N

The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N

Learn more here: brainly.com/question/25892144

4 0
2 years ago
A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 5100 MJ with respect to Eart
serious [3.7K]

To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,

E_{initial} = E_{final}

PE_{initial}+KE_{initial} = PE_{final}+KE_{final}

Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that

KE_{final} = (PE_{initial}+KE_{initial} )-PE_{final}

KE_{final} = (5100+4200)-5700

KE_{final} = 3600MJ

Therefore the final kinetic energy is 3600MJ

5 0
3 years ago
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