The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.
To find the answer, we need to know about the tension.
<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>
- Let's draw the free body diagram of the system using the given data.
- From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
- For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

- We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

- To find Ny, we need to find the tension T.
- For this, we can equate the net horizontal force.

- Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,

- Thus, the magnitude of the force that the beam exerts on the hi.nge will be,

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.
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Answer:
The answer is "
".
Explanation:
Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

Potential energy shifts:


Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.



This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.
Well, if you're using the law to work with periods of Earth satellites,
then the most convenient unit is going to be 'hours' for the largest
orbits, or 'minutes' for the LEOs.
But if you're using it to work with periods of planets, asteroids, or
comets, then you'd be working in days or years.
The answer is: "
44
km " ;
or; write as: "
44.333 km " .
___________________________________________________________Explanation:___________________________________________________________(70 km + 63 km) ÷ (2 + 1 ) = 133 km ÷ 3 = "
44
km " ;
or; write as: "
44.333 km " .
___________________________________________________________
I think your answer would be DDT, but don't hold me to it