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hoa [83]
4 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]4 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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3 years ago
What is the difference between speed and velocity?
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8 0
3 years ago
Amazon rectangular bar of low carbon steel A-36 is exposed to an axial strees of 150 MPa. What is the original length of the bar
kolbaska11 [484]

Answer:

1.8m

Explanation:

Let the Elastics of the steel ASTM-36 E = 200000 MPa

The strain of the bar when subjected to 150 MPa is

\epsilon = \frac{\sigma}{E} = \frac{150}{200000} = 0.00075

Therefore, if the bar elongates by 1.35 mm, then the original length L would be:

\epsilon = \frac{\Delta L}{L}

L = \frac{\Delta L}{\epsilon} = \frac{1.35}{0.00075} = 1800 mm or 1.8m

5 0
3 years ago
Consider three force vectors Fi with magni- tude 53 N and direction 116º, F2 with mag- nitude 57 N and direction 217°, and F3 wi
Flauer [41]

Answer:

a. Fnet =37.67N

b. The direction = 133.4 from the x axis counter clockwise.

c. Option 2

Explanation:

Given that F1 is 53N at 116°, then it will be at a direction of 116-90=26° in the second quadrant.

Given that F2 is 57N at 116°, then it will be at a direction of 217-180=37° in the third quadrant..

Given that F1 is 71N at 20°, then it is in the first quadrant.

a. Fnet= F1+F2+F3

Fnet= -F1sin26i+F2cos26j-F2cos37i-F2sin37j+F3cos20i+F3sin20j

Fnet= 53sin26i+53cos26j-57cos37i-57sin37j+71cos20i+71sin20j

Resolving the vectors into x and y components.

Fnet= -2.04i+37.62j

Magnitude of the vector

Fnet= √((-2.04)^2+(37.62)^2)

Fnet= 37.67N

Fnet is approximately 38N.

b. Direction of the Fnet.

Angle=arctan(y/x)

Angle=arctan(-37.61/2.04)

Angle= -43.37°

The angle is in the negative x axis and positive y axis.

Then the direction becomes 180-43.37

Therefore, the direction of the net force is 133.37°.

c. The instantaneous velocity of a body is always in the direction of the net force at that instant. Option 2 is correct.

Fnet=ma

Fnet= mv/t

So the velocity is in the direction of the Fnet.

3 0
3 years ago
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