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Andrews [41]
4 years ago
7

A 11 g plastic ball is moving to the left at 29 m/s. How much work must be done on the ball to cause it to move to the right at

29 m/s ? Express your answer using two significant figures.
Physics
1 answer:
Rasek [7]4 years ago
8 0

Answer:

4.25 J

Explanation:

Given that

mass of plastic ball = 11 g

Mass of plastic ball = 0.011 kg

velocity of ball = 29 m/s

We know that from work power energy theorem

W_{all}=Change\ in\ kinetic\ energy\ of\ system

We know that kinetic energy of moving mass given as

KE=\dfrac{1}{2}mv^2

Now by pitting the values

KE=\dfrac{1}{2}mv^2

KE=\dfrac{1}{2}\times 0.011\times 29^2

KE= 4.25 J

So the work done on the ball is 4.25 J

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The hydraulic oil in a car lift has a density of 8.53 x 102 kg/m3. The weight of the input piston is negligible. The radii of th
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(a) the input force is 36.56 N

(b) the input force is 37.49 N

Explanation:

Given;

density of hydraulic oil, ρ =  8.53 x 10² kg/m³

radius of plunger, r₁ = 0.135 m

radius of piston, r₂ = 5.43 x 10⁻³ m

Part (a) The input force needed to support 22600-N weight, when the bottom surfaces of the piston and plunger are at the same level;

P =\frac{F}{A}

Where;

P is pressure

F is force

A is circular area = πr²

\frac{F_1}{A_1} =\frac{F_2}{A_2} \\\\F_2 = \frac{F_1*A_2}{A_1} =\frac{F_1* \pi r_2^2}{\pi r_1^2} = \frac{F_1*  r_2^2}{ r_1^2} \\\\F_2 = \frac{22600*(5.43*10^{-3})^2 }{(0.135)^2}\\\\F_2 = 36.56 \ N

Part (b) The input force needed to support 22600-N weight, when the  bottom surface of the output plunger is 1.20 m above that of the input plunger

P_2 = P_1 + \rho gh

But, F = PA  and  A = πr²

F_2 = F_1(\frac{A_2}{A_1} ) + \rho gh*A_2\\\\F_2 = F_1(\frac{r_2^2}{r_1^2} )+\rho gh(\pi r_2^2)\\\\F_2 = 22600(\frac{5.43*10^{-3}}{0.135})^2 \ + 853*9.8*1.2*\pi (5.43*10^{-3})^2\\\\F_2=36.56 + 0.93\\\\F_2 = 37.49 \ N

4 0
3 years ago
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