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3 years ago

An astronaut is said to be weightless when he/she travels in a satellite. Does it mean that the earth does not attract him/her

1 answer:

7 0

First of all, what a weight is? weight is m × g.. and when a person is in space then the force of gravity isn't acting on him or her, so yeah it will not have weight, but remember gravity is a very very low, now if a body has more mass, more gravity will act on it right? just imagine the mass of a person and that of earth.. now yes, earth will do exert a force on the astronaut but is negiligable.

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A 495-kg dragster accelerates from rest to a final speed of 105 m/s in 395 m, during which it encounters an average frictional f

drek231 [11]

Answer:

Explanation:

According to energy conservation which states that the workdone is equal to change in the system

Workdone = change in kinetic energy + (frictional force * distance)

Workdone = ΔK + fd

Workdone = kf-Ki + fd

Workdone = = 1/2(m(v-u)^2) + fd

Given

Mass m = 495kg

final velocity v = 105m/s

initial velocity = 0m/s

Force f= 1400N

distance d = 395m

Substitute

Workdone = 1/2(495(105-0)^2) + 1400(395)

Workdone = 2,728,687.5+553000

Workdone = 3,281,687.5 Joules

Time = 8.2secs

Power output = Workdone/Time

Power output = 3,281,687.5/8.2

Power output = 885,766.768

Power output = 8.858 * 10^5 watts

3 0

4 years ago

A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will

alukav5142 [94]

Answer:

The change in momentum is $\Delta p = kg \cdot m/s$

Explanation:

From the question we are told that

The mass of the probe is $m = 170 kg$

The location of the prob at time t = 22.9 s is $A =$

The momentum at time t = 22.9 s is $p = < 51000, -7000, 0> kg m/s$

The net force on the probe is $F = N$

Generally the change in momentum is mathematically represented as

$\Delta p = F * \Delta t$

The initial time is 22.6 s

The final time is 22.9 s

Substituting values

$\Delta p = * (22.9 - 22.6)$

$\Delta p = * (0.3)$

$\Delta p = kg \cdot m/s$

6 0

3 years ago

The average speed of an object is found by dividing the _________ by the time of travel.

nadya68 [22]

distance travelled by the mobile

6 0

3 years ago

Read 2 more answers

A speedboat initially at rest accelerates at 4.0 m/s (squared) for 7.0 s. How far does the speedboat move in 7.0 s?

Ulleksa [173]

I think is 3 becuase when he star at 4

4 0

3 years ago

8. A 30-kg box is sliding down a frictionless plane that is sloped at 24º. Assuming the object starts at rest,

Katena32 [7]

The net force on the box parallel to the plane is

∑ F[para] = mg sin(24°) = ma

where mg is the weight of the box, so mg sin(24°) is the magnitude of the component of its weight acting parallel to the surface, and a is the box's acceleration.

Solve for a :

g sin(24°) = a ≈ 3.99 m/s²

The box starts at rest, so after 7.0 s it attains a speed of

(3.99 m/s²) (7.0 s) ≈ 28 m/s

6 0

2 years ago