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Y_Kistochka [10]
2 years ago
6

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

enough above the earth's surface that they have escaped the pull of earth's gravity. the space shuttle has anti-gravity devices on it that allow the astronauts to float inside it. the astronauts are falling at the same rate as the space shuttle as it orbits around earth. orbiting objects do not experience any gravity?
Physics
1 answer:
Maslowich2 years ago
7 0

If we see the forces on the astronauts then there is only one force on each astronaut while they are in air.

This force is due to gravity of earth.

Now while the astronauts are in air and doing some activities then the net force on them is counterbalanced on them by centrifugal force.

so we can say the rate of fall of astronauts due to gravity is at same rate as the orbiting rate of the space shuttle.

this is given by

w^2R = g

so all astronauts will experience the situation of free fall.

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A point charge with charge q1 is held stationary at the origin. A second point charge with charge q2 moves from the point (x1, 0
Scilla [17]

Answer:

W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})

Explanation:

Position of charge q₁ is (0,0)

Position of charge q₂ is (x₁,0)

So, the electric potential energy between the charges is given by :

U_1=k\dfrac{q_1q_2}{x_1}

Now the position of charge q₂ has been changes from (x₁,0) to (x₂,y₂). Now, electric potential energy between the charges is :

U_2=k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}}

We know form the work energy theorem that, the change in potential energy is equal to the work done. Mathematically, it is given by :

W=-\Delta U

W=-(U_2-U_1)

W=(U_1-U_2)

W=(k\dfrac{q_1q_2}{x_1}-k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}})

W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})

Hence, the work done by the electrostatic force on the moving point charge is kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}}). Hence, this is the required solution.

3 0
2 years ago
What best describes myotibrils
krok68 [10]
Myofibrils are composed of long proteins such as actin, myosin, and titin, and other proteins that hold them together. These proteins are organized into thin filaments and thick filaments, which repeat along the length of the myofibril in sections called sarcomeres. Muscles contract by sliding the thin (actin) and thick (myosin) filaments along each other.
8 0
2 years ago
Identical guns fire identical bullets horizontally at the same speed from the same height above level planes, one on the Earth a
Natasha_Volkova [10]

Answer:

I. The horizontal distance traveled by the bullet is greater for the Moon.

II. The flight time is less for the bullet on the Earth.

Explanation:

Horizontal distance depends on the initial speed, height and gravity. Bullets have the same initial speed and are shot from the same height. In these conditions horizontal distance only depends on gravity, which is inversely proportional. Therefore, the less gravity the greater the horizontal distance. Gravity slows bullet and causes its impact on the ground. Since gravity is greater in Earth, the bullet hits faster on the earth.

3 0
1 year ago
Read 2 more answers
12*8A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 13 feet. The ball is started in
max2010maxim [7]

Answer:

See attached pictures.

Explanation:

See attachments for explanation.

6 0
2 years ago
A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, take
adell [148]

Answer:

F_{x}=2.31N to the right.

F_{y}=2.1N to in the upwards direction.

Explanation:

In order to solve this problem, we must first start by drawing a diagram of the situation. (See attached diagram).

So, remember that a force is determined by multiplying the mass of the parcticle by its acceleration:

F=ma

so in order to find the components of the force, we need to start by finding its acceleration.

Acceleration is found by using the following formula:

a=\frac{V_{f}-V{0}}{t}

so we can subtract the two vectors, like this:

a=\frac{(6.00i+4.0j)m/s-1.60i m/s}{8s}

which yields:

a=\frac{(4.4i+4.0j)m/s}{8s}

or:

a=(0.55i + 0.5j) m/s^{2}

so now I can find the components of the force:

F=(4.2kg)(0.55i + 0.5j) m/s^{2}

which yields:

F=(2.31i+2.1j)N

so the components of the force are:

F_{x}=2.31N to the right.

F_{y}=2.1N to in the upwards direction.

6 0
1 year ago
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