a centrifugal clutch works, as the name suggests, through centrifugal force. ... The rotation of the hub forces the shoes or flyweights outwards until they come into contact with the clutch drum, the friction material transmits the torque from the flyweights to the drum. The drive is then connected
The speed of light is 299,792,458 meters per second in vacuum,
somewhat less in any material. How much less depends on the material.
Answer:
25kg
Explanation:
The formula for acceleration due to gravity is:

Here, G represents a constant and M represents the mass of the object. We don't need to know the actual values to solve this, then: We can just multiply by the amount they are multiplied relative to earth.

Multiplying this by the weight of the person on Earth of 40kg, you get 0.625*40=25kg.
Hope this helps!
Answer:

Explanation:
Given data:
v = 220 rms
power factor = 0.65
P = 1250 W
New power factor is 0.9 lag
we knwo that

s = 1923.09 < 49.65^o
s = [1250 + 1461 j] vA
![P.F new = cos [tan^{-1} \frac{Q_{new}}{P}]](https://tex.z-dn.net/?f=P.F%20new%20%3D%20cos%20%5Btan%5E%7B-1%7D%20%5Cfrac%7BQ_%7Bnew%7D%7D%7BP%7D%5D)
solving for 
![Q_{new} = P tan [cos^{-1} P.F new]](https://tex.z-dn.net/?f=Q_%7Bnew%7D%20%3D%20P%20tan%20%5Bcos%5E%7B-1%7D%20P.F%20new%5D)
![Q_{new} = 1250 [tan[cos^{-1}0.9]]](https://tex.z-dn.net/?f=Q_%7Bnew%7D%20%3D%201250%20%5Btan%5Bcos%5E%7B-1%7D0.9%5D%5D)






Faraday

To gather light from a far-off star, a concave mirror with a radius of curvature of 1.0 m is utilized. Most likely 0.25 meters separate the mirror from the star's picture.
A concave mirror is a kind of spherical mirror in which the reflecting surface is the inside radius of curvature of the sphere, or in other words, the reflecting surface appears to be far from the incident light source. Is called concave mirror.
The reciprocal of the curvature is known as the radius of curvature, or R. It is equal to the circumference of the circular arc that, at that location, most closely resembles the curve. The circle's radius that best matches a normal section is known as the radius of curvature for surfaces.
Learn more about concave mirror here
brainly.com/question/3555871
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