The value of the force, F₀, at equilibrium is equal to the horizontal
component of the tension in string 2.
Response:
- The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>
<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>
Given:
The weight of the rod = The sum of the vertical forces in the strings
Therefore;
M·g = T₂·cos(37°) + T₁
The weight of the rod is at the middle.
Taking moment about point (2) gives;
M·g × L = T₁ × 2·L
Therefore;

Which gives;


F₀ = T₂·sin(37°)
Which gives;

<u />
Learn more about equilibrium of forces here:
brainly.com/question/6995192
Answer:
b.only when the current in the first coil changes.
Explanation:
An induced current flow in the second coil only when there is a change in current in the first cool. A steady current will produce no change in flux (due to magnetic effect of a current) by the first coil, and according to Faraday, induced current is only produced when there is a change in flux linkage.
Answer:
The work required is -515,872.5 J
Explanation:
Work is defined in physics as the force that is applied to a body to move it from one point to another.
The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.
Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).
The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:


In this case:
- W=?
- m= 2,145 kg
- v2= 12

- v1= 25

Replacing:

W= -515,872.5 J
<u><em>The work required is -515,872.5 J</em></u>
Answer:
Here are 5:
Distance from source to receiver
Wind speed and direction
Wind gradients
Temperature gradients
Atmospheric attenuation
and there are many more...
Hope that was helpful.Thank you!!!