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Ksivusya [100]
4 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]4 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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Read 2 more answers
1. A hydraulic lift is to be used to lift a truck masses 5000 kg. If the diameter of the
zheka24 [161]

Answer:

a)   F = 4.9 10⁴ N,  b)   F₁ = 122.5 N

Explanation:

To solve this problem we use that the pressure is transmitted throughout the entire fluid, being the same for the same height

1) pressure is defined by the relation

           P = F / A

to lift the weight of the truck the force of the piston must be equal to the weight of the truck

          ∑F = 0

          F-W = 0

          F = W = mg

          F = 5000 9.8

          F = 4.9 10⁴ N

the area of ​​the pisto is

          A = pi r²

          A = pi d² / 4

          A = pi 1 ^ 2/4

          A = 0.7854 m²

pressure is

          P = 4.9 104 / 0.7854

          P = 3.85 104 Pa

2) Let's find a point with the same height on the two pistons, the pressure is the same

          \frac{F_1}{A_1} = \frac{F_2}{A_2}

where subscript 1 is for the small piston and subscript 2 is for the large piston

          F₁ = \frac{A_1}{A_2} \ F_2

the force applied must be equal to the weight of the truck

          F₁ = ( \frac{d_1}{d_2} )^2\  m g

          F₁ = (0.05 / 1) ² 5000 9.8

          F₁ = 122.5 N

7 0
3 years ago
A stubborn, 100 kgkg mule sits down and refuses to move. To drag the mule to the barn, the exasperated farmer ties a rope around
jolli1 [7]

Answer:

No, the farmer is not able to move the mule.

Explanation:

Mass =100 kg

Force=F=800 N

The coefficient between the mule and the ground=\mu_s=0.8

\mu_k=0.5

Static friction force,f=\mu N

Normal force=N=mg

Static friction force,f=\mu_s mg=0.8\times 100\times 9.8=784 N

Using g=9.8m/s^2

F<f

Static friction force is greater than applied force.

Therefore , the farmer is not able to move the mule.

4 0
3 years ago
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