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Ksivusya [100]
3 years ago
10

A ball is thrown vertically downwards at speed vo from height h. Draw velocity vs. time & acceleration vs. time graphs. In t

erms of only the givens vo and h, derive expressions for the final speed of the ball and the elapsed time of flight.

Physics
1 answer:
yuradex [85]3 years ago
4 0

Answer:

Explanation:

let the ball is thrown vertically downwards with velocity u.

So, initial velocity, = - u (downwards)

acceleration = - g (downwards)

let the velocity is v after time t.

use first equation of motion

v = u + at

- v = - u - gt

v = u + gt

So, it is a straight line having slope g and y intersept is u.

The graph I shows the velocity - time graph.

Now the value of acceleration remains constant and it is equal to - 9.8 m/s^2.

So, acceleration time graph is a starigh line parallel to time axis having slope zero.

the graph II shows the acceleration - time graph.

Use III equation of motion to find the final speed in terms height.

v^{2}=u^{2}+2gh

And the time is

v = u + gt

t=\frac{v-u}{g}

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d) 1.44m

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applying the second law

-1.25m=m.a\\a=-1.25m/s^2

given the acceleration we can calculate the distances traveled before stopping:

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8 0
3 years ago
Two upwards forces act: one of 34N which is 3.5m to the left of the axle, the other is 25N and is 2.6m to the right of the axle.
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Answer:

See explanation below

Explanation:

In this case, you want to know if you put an object between these forces, which direction would go.

To know this, we need to calculate the moment of an object, which is defined as the product of a force and it's distance. In other words:

M = F * d   (1)

And, in order to reach equilibrium the force will exert a direction in clockwise or anticlosewise, and these moments, should be even:

anticlockwise moment = clockwise moment.

The clockwise would be the forces to the right, and anticlock would the only force to the left of the axle.

Clockwise moment = (10 * 0.8) + (25 * 2.6) = 73 Ns

Anticlockwise moment = 34 * 3.5 = 119 Ns.

As we can see, the moment in the anticlockwise is higher than the actual clockwise moment, therefore, we can assume that the object will move anticlockwise, or simply move to the left.

Hope this helps

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Current=I

\boxed{\sf P=I^2R}

\\ \sf\longmapsto I^2=\dfrac{P}{R}

\\ \sf\longmapsto I^2=\dfrac{9.28}{210}

\\ \sf\longmapsto I^2\approx0.04

\\ \sf\longmapsto I\approx\sqrt{0.04}

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