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3 years ago

10

in hydraulic pressure the area of the larger piston is half of the smaller piston.if the force of 4000N is applied on the smalle

r piston.what is the distance covered by the train ?

1 answer:

irina1246 [14]3 years ago

7 0

Answer:

The distance moved by the train is half the distance moved by the force

Explanation:

Given that the force applied to the smaller piston, F₂ = 4,000 N

The pressure in the area of the larger piston, A₁ = 2 × The pressure in the smaller piston, A₂

∴ A₁ = 2 × A₂

We have;

$\dfrac{A_1}{A_2} = \dfrac{2\cdot A_2 }{A_2} =2 = \dfrac{F_1}{F_2}$

F₁ = 2·F₂ = 2 × 4,000 N = 8,000 N

F₁·d₁ = F₂·d₂

∴ 8,000 N ×d₁ = 4,000 N × d₂

d₁ = 1/2·d₂

∴ The distance moved by the train is half the distance moved by the force

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In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
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Using this law, we can answer all the three questions of the problem.

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And then, the charge at time $t=1.95 \tau$ is equal to
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<em>b) 61.908 J</em>

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Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

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$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

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and mass 11 kg

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we know that v = ωr

where v is the linear speed = 35 m/s

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therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

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