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1 year ago

An astronaut leaves a spaceship that is moving through free space. Will the spaceship move off and leave the astronaut behind?

1 answer:

9 0

When an astronaut leaves a spaceship that is moving through free space, the spaceship does not move off and leave the astronaut behind.

This is because when the astronaut is inside the spaceship, which is moving in free space, the astronaut also has the same velocity as the spaceship. Hence, when he leaves the spaceship, there is no relative velocity between the spaceship and the astronaut. This is the reason why the spaceship will not move off and leave the astronaut behind.

A spaceship or a spacecraft is an especially designed vehicle that is used to travel to outer space. It can be used for space exploration, transportation of humans and cargo, and meteorological purposes.

An astronaut is a person who has undergone specific training to travel to space. The duties of an astronaut aboard a spaceship can vary. Usually, a pilot and a commander work together to lead the mission. Other positions could be science pilot, mission expert, payload commander, and flight engineer.

Refer to more about astronaut here

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Why do astronomers use spectroscopes to analyze light from distant objects?

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3 years ago

A sample of a gas in a rigid container has an initial pressure of 5 atm at a temperature of 254.5 k. The temperature is decrease

skelet666 [1.2K]

The gas is in a rigid container: this means that its volume remains constant. Therefore, we can use Gay-Lussac law, which states that for a gas at constant volume, the pressure is directly proportional to the temperature. The law can be written as follows:

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$

Where P1=5 atm is the initial pressure, T1=254.5 K is the initial temperature, P2 is the new pressure and T2=101.8 K is the new temperature. Re-arranging the equation and using the data of the problem, we can find P2:

$P_2 = T_2 \frac{P_1}{T_1}=(101.8 K) \frac{5 atm}{254.5 K}=2 atm$

So, the new pressure is 2 atm.

7 0

3 years ago

If a 1.00 kg body has an acceleration of 2.44 m/s2 at 53Â° to the positive direction of the x axis, then what are (a) the x comp

Ilia_Sergeevich [38]

(a) Fx = 1.464 N

(b) Fy = 1.952 N

Given

Mass = 1kg

Acceleration = 2.44 m/s2

Angle with positive X axis = 53°

As we know

F = ma

By substituting value

F= 1×2.44 N

F= 2.44 N

(a) Component of force in X direction

Fx = F Cosθ

Fx = 2.44 Cos(53°)

Fx = 2.44 × 0.60 = 1.464 N

(b) Component of force in Y direction

Fy = F Sinθ

Fy = 2.44 Sin(53°) = 2.44 × 0.80 = 1.952 N

F(x, y) = 1.464 i + 1.952 j

Thus we got net force.

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6 0

2 years ago

You have a pick-up truck that weighed 4,000 pounds when it was new. you are modifying it to increase its ground clearance. when

jok3333 [9.3K]

U have to *modify it to increase its ground clearance*

4 0

3 years ago

Read 2 more answers

A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .

rjkz [21]

a) 2.64 s

We can solve this part of the problem by using the following SUVAT equation:

$s=ut+\frac{1}{2}at^2$

where

s is the displacement of the stone

u is the initial velocity

t is the time

a is the acceleration

We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:

- s (displacement) negative, since it is downward: so s = -75.0 m

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)

Substituting into the equation,

$-75.0 = 15.5 t -4.9t^2\\4.9t^2-15.5t-75.0 = 0$

Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.

b) 10.4 m/s

The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:

$v=u+at$

where again, we must be careful to the signs of the various quantities:

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2

Substituting t = 2.64 s, we find the final velocity of the stone:

$v = 15.5 +(-9.8)(2.64)=-10.4 m/s$

where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.

c) 4.11 s

In this case, we can use again the equation:

$s=ut+\frac{1}{2}at^2$

where

s is the displacement of the package

u is the initial velocity

t is the time

a is the acceleration

We have:

s = -105 m (vertical displacement of the package, downward so negative)

u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)

a = g = -9.8 m/s^2

Substituting into the equation,

$-105 = 5.40 t -4.9t^2\\4.9t^2 -5.40 t-105=0$

Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after

t = 4.11 seconds.

5 0

3 years ago

Read 2 more answers