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3 years ago

6

Calculate the applied force to the piston with a 12cm radius required to elevate a weight of 2.0X104N by the piston with a 36cm

radius in a hydraulic lift.
2.9 × 103 N

5.0 × 103 N

6.7 × 103 N

2.2 × 103 N

1 answer:

Alex3 years ago

6 0

Answer:

Option D is the correct answer.

Explanation:

Refer the figure given.

By Pascal's principle we have

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

F2 = 2 x 10⁴ N

$A_1=\frac{\pi\times (12\times 10^{-3})^2}{4}=1.13\times 10^{-4}m^2\\\\A_2=\frac{\pi\times (36\times 10^{-3})^2}{4}=1.02\times 10^{-3}m^2$

Substituting

$\frac{F_1}{1.13\times 10^{-4}}=\frac{2\times 10^4}{1.02\times 10^{-3}}\\\\F_1=2.22\times 10^3N$

Option D is the correct answer.

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Given :

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g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

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Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

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$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

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substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

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And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

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So

$Q = 0.3 + 0.18$

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$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

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substituting value

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And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

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From the law of energy conservation

$Q = Q_f$

So

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

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