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Stels [109]
3 years ago
10

Why to astronauts appear weightless while they are filmed performing activities inside the orbiting space shuttle?

Physics
2 answers:
vagabundo [1.1K]3 years ago
8 0
<h2>Answer: The astronauts are falling at the same rate as the space shuttle as it orbits around earth</h2>

The astronauts seem to float because they are in free fall just like the spacecraft.

However, although they are constantly falling on the Earth, they do not fall because the ship orbits at a sufficient speed (in the same direction of rotation of the Earth) so that the centrifugal force is balanced with the Earth's gravitational pull.

In other words:

The spaccraft and the astronauts are in free fall but the Earth's surface will never be reached as long as they does not decrease the speed.

Then, as they accelerate toward Earth (regardless of their mass), it curves beneath them and never comes close.

That's why astronauts, having the same acceleration as the spacecraft, feel weightless and see themselves floating.

Alexeev081 [22]3 years ago
5 0

Answer:

an astronaut is under the gravitational force of any object nearby that has mass.

Explanation:

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What is time taken by the stone ?
SSSSS [86.1K]

The time taken by the stone to hit the ground would be 5.12 seconds.

<h3>What are the three equations of motion?</h3>

There are three equations of motion given by  Newton

The first equation is given as follows

v = u + at

the second equation is given as follows

S = ut + 1/2×a×t²

the third equation is given as follows

v² - u² = 2×a×s

Keep in mind that these calculations only apply to uniform acceleration.

As given in the problem, a stone is dropped from the helicopter which is ascending at the speed of 19.6 m/s

height(S) = 156.8 meters

initial velocity(u) = -19.6 m/s

acceleration(a) = 9.81 m/s²

By using the second equation of motion given by newton

S = ut + 1/2at²

S = 156.8m ,u= -19.6 m/s , a= 9.81 m/s² and t =? seconds

156.8= -19.6t + 9.81t²

t = 5.12 seconds

Thus, the time taken by the stone to hit the ground would be 5.12 seconds.

Learn more about equations of motion from here,

brainly.com/question/5955789

#SPJ1

8 0
2 years ago
A 4,000 kg truck is moving at +10 m/s hits a 1500 kg parked car which moves off at +10 m/s. What is the velocity of the truck
Igoryamba

Answer:

10m/s

Explanation:

Using the law of conservation of momentums

M1u1+m2u2 = (m1+m2)v

Substitute.

4000(10)+1500(10) = (4000+1500)v

40,000+15,000 = 5,500v

55000 = 5500v

v = 55000/5500

v= 10m/s

Hence the velocity of the truck after Collision is 10m/s

6 0
3 years ago
How do defy gravity?
kaheart [24]
Exert force upward.
Like when you pick something up from the floor, or walk up the stairs.
6 0
4 years ago
Can you answer this math homework? Please!
kap26 [50]

\large \mathfrak{Solution : }

9. An object which is in circular motion (moving along a circle) is said to be accelerating because it changes it's direction constantly even if it is moving with a constant speed. cuz acceleration is change in either magnitude or direction of an object with respect to time.

therefore, it's still acceleration as change in direction with time.

10. Average speed of an object can be calculated by dividing the total distance covered by an object by time taken to cover that distance.

i.e

  • \boxed{speed =  \dfrac{distance}{time} }

it can be re- arranged to find the distance as :

  • \boxed{distance = speed \times time}

  • time =  \dfrac{distance}{speed}

11. speed = 20 m/s : conversion into km/h

distance covered : 4 km = 4000 m

  • time =  \dfrac{distance}{speed}

  • t =  \dfrac{4000}{20}

  • t =  200 \:  \: sec

time taken = 200 seconds

12. let's use the first equation of motion to find the acceleration :

  • v = u + at

  • 50 = 80 + 120a

  • 50 - 80 = 120a

  • a =  \dfrac{ - 30}{120}

  • a = -   \dfrac{1}{4} m/s {}^{2}

  • - 0.25 \:  \: m/s {}^{2}
3 0
3 years ago
Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i
Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

V = 4\frac{kQ}{r}

here

r = distance of corner of the square from it center

r = \frac{a}{\sqrt2}

r = \frac{10nm}{\sqrt2} = 7.07 nm

Q = e = -1.6 * 10^{-19} C

now the net potential is given as

V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}

V = 0.815 V

now potential energy of alpha particle at this position

U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J

Now at the mid point of one of the side

Electrostatic potential is given as

V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}

here we know that

r_1 = \frac{a}{2} = 5 nm

r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}

r_2 = 11.2 nm

now potential is given as

V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}

V = -0.576 - 0.257 = -0.833 V

now final potential energy is given as

U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J

Now work done in this process is given as

W = U_f - U_i

W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}

W = -7 * 10^{-22} J

8 0
3 years ago
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