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hoa [83]
3 years ago
14

An astronaut tightens a bolt on a satellite in orbit. He rotates in a direction opposite to that of the bolt, and the satellite

rotates in the same direction as the bolt. Explain why. If a handhold is available on the satellite, can this counter-rotation be prevented
Physics
1 answer:
statuscvo [17]3 years ago
4 0

Answer: yes it can be prevented

Explanation:

The sensation of weightlessness that astronauts experience seems to make their tasks almost effortless. However, as Newton's third law of motion suggests, working in space can be physically demanding. 

As he tightens the bolt, he is rotating in the direction opposite to the bolt

It is possible if the handhold is designed in three dimensional motion where the astronaut motion will be the uplimb motion with the mass centre of hand move along circular helix trajectory

Angular momentum is conserved astronaut motion is conserved when net external torque is Zero.

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A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha
USPshnik [31]

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

\Phi_E=\frac{Q}{\epsilon_o}

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C

Finally, you obtain for E:

E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

3 0
3 years ago
Answers please!!!!!!!!!!!!!!!!
kaheart [24]
I only know #2 and #4.
2.) cells
3.) cells,  life , existing

Sorry that i dont know the rest but i took a test on this not to long ago, and i tend to forget stuff once i take a test on it.


3 0
4 years ago
Which explanation is based on scientific evidence?​
Yuki888 [10]

Answer:

Scientific evidence is evidence that serves to either support or counter a scientific theory or hypothesis. Such evidence is expected to be empirical evidence and interpretable in accordance with scientific method.

5 0
3 years ago
A ball is thrown up at a speed of 20 m/s.
ioda

Given:

(Initial velocity)u=20 m/s

At the maximum height the final velocity of the ball is 0.

Also since it is a free falling object the acceleration acting on the ball is due to gravity g.

Thus a=- 9.8 m/s^2

Now consider the equation

v^2-u^2= 2as

Where v is the final velocity which is measured in m/s

Where u is the initial velocity which is measured in m/s

a is the acceleration due to gravity measured in m/s^2

s is the displacement of the ball in this case it is the maximum height attained by the ball which is measured in m.

Substituting the given values in the above formula we get

0-(20x20)= 2 x- 9.8 x s

s= 400/19.6= 20.41m

Thus the maximum height attained is 20.41 m by the ball

6 0
3 years ago
What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL
lozanna [386]

Answer:

\displaystyle \Delta x=1.74\ m

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

\displaystyle PE = \frac{1}{2}k(\Delta x)^2

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

\displaystyle \Delta x=\swrt{\frac{2PE}{k}}

Substituting:

\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}

Calculating:

\displaystyle \Delta x=\sqrt{3.0175}

\boxed{\displaystyle \Delta x=1.74\ m}

6 0
3 years ago
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