How to calculate moments with 3 separate weights of different amounts at different points?

1 answer:

6 0

I don't completely understand your drawing, although I can see that you certainly
did put a lot of effort into making it. But calculating the moment is easy, and we
can get along without the drawing.

Each separate weight has a 'moment'.
The moment of each weight is:

(the weight of it) x (its distance from the pivot/fulcrum) .

That's all there is to a 'moment'.

The lever (or the see-saw) is balanced when (the sum of all the moments
on one side) is equal to (the sum of the moments on the other side).

That's why when you're on the see-saw with a little kid, the little kid has to sit
farther away from the pivot than you do. The kid has less weight than you do,
so he needs more distance in order for his moment to be equal to yours.

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Diego is playing basketball. While running at 7 km/h toward Gino, he passes the ball to Gino horizontally. The ball travels at 2

enot [183]

TheThe relative velocity of Deigo will be 27km/h .

<h3>What is relative Velocity? </h3>

The relative velocity is the velocity of an object with respect to another observer. It is the time rate of relative position of one object with respect to another object .

Vab = Va-Vb

Where a is object and b is observer .

We have given here the velocity of ball 20km/ h

and the another velocity is 7km/ h

Let us assume that Diego is also moving in the direction where ball is moving so the relative velocity will be

V = 20+7=27km/ h

When both observer and object is moving in same direction than there sum of velocity will be relative velocity ,if they are moving opposite than subtraction.

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2 years ago

A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.

erastova [34]

Complete Question

A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.

(a) How much heat transfer is needed each second to raise the water temperature from 35.0°C to 100°C, boil it, and then raise the resulting steam from 100°C to 450°C? Specific heat of water is 4184 J/(kg · °C), the latent heat of vaporization of water is 2256 kJ/kg, and the specific heat of steam is 1520 J/(kg · °C).

(b) How much power is needed in megawatts? (Note: In real power plants, this process occurs under high pressure, which alters the boiling point. The results of this problem are only approximate.)

Answer:

The heat transferred is $Q = 5.866 * 10^9 J$