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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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1500000 Pa

Explanation:

The formula for pressure is force per unit area.

P=F/A where  F is force and A is area

Given that ;

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P= 1471500

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3 years ago
At one instant, a 17.0-kg sled is moving over a horizontal surface of snow at 4.10 m/s. After 6.15 s has elapsed, the sled stops
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Answer:

force = 11.33 kg-m/s^{2}

Explanation:

given data:

sled mass = 17.0 kg

inital velocity (U) = 4.10 m/s

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final velocity (V) = 0

final momentum P2 = 0

Initial momentum of sledge is

P_{1}=mU

P_{1}= 17.0 * 4.10 = 69.7 kg- m/s

from newton second law of motion

F=\frac{\Delta P}{\Delta t}

F = \frac{P_{1}-P_{2}}{T}

Kgm/s^2

F = \frac{69.7-0}{6.15}= 11.33[tex]kg-m/s^{2}[/tex]

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Explain how surface and subsurface events are integral parts of the rock cycle.
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Example of surface events are erosion and weathering. Erosion is the carrying of a particle from one place to the other and weathering is the breaking down of particles. These processes help in rock formation because this allows physical changes (grouping together or breaking down) on a certain substance. Subsurface events are those which happened underground such as the flow of underground water which subsequently allow the deposition of minerals, etc. 
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Answer:

Mass of the climber = 69.38 kg

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Young's modulus, Y = 0.37 x 10¹⁰ N/m²

Area

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Length, L = 15 m

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Substituting

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Mass of the climber = 69.38 kg

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