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Alex_Xolod [135]

3 years ago

13

n input force is applied to a 2 cm diameter piston. The piston compresses oil and is connected to a 4 cm diameter output piston.

The force on the output piston is ________ the force on the input piston.

1 answer:

boyakko [2]3 years ago

6 0

Answer:

The force on the output piston is 4 times of the force on the input piston.

Explanation:

Given that,

The diameter of the input piston, $d_1=2\ cm$

The radius of the input piston, $r_1=1\ cm$

The diameter of the output piston, $d_2=4\ cm$

The radius of the output piston, $r_2=2\ cm$

According to Pascal's law, the pressure at smaller piston is equal to the pressure at larger piston such that,

$\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}\\\\\dfrac{F_2}{F_1}=\dfrac{A_2}{A_1}\\\\\dfrac{F_2}{F_1}=\dfrac{\pi r_2^2}{\pi r_1^2}\\\\\dfrac{F_2}{F_1}=\dfrac{r_2^2}{r_1^2}\\\\\dfrac{F_2}{F_1}=\dfrac{2^2}{1^2}\\\\\dfrac{F_2}{F_1}=\dfrac{4}{1^2}\\\\F_2=4\times F_1$

So, the force on the output piston is 4 times of the force on the input piston.

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A 4x100 relay is where 4 people run 100 meters and a 4x400 relay is where 4 people run 400 meters

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3 years ago

Mademuasel [1]

According to Newton's Second Law of Motion :

The Force acting on an Object is equal to Product of Mass of the Object and Acceleration produced due to the Force.

$:\implies$ Force acting = Mass of the Object × Acceleration

Given : Force = 50 newton and Mass of the Object = 10 kg

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A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

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a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

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The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

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