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stealth61 [152]

2 years ago

15

A piston in an internal combustion engine applies torque during 150° of each full rotation of the crankshaft to which it is conn

ected. Suppose an 8-cylinder engine (8 such pistons) is running with its crankshaft turning at 2500 rpm and produces 410J of work each second. What is the torque applied to the crank shaft by one of the engine’s eight pistons?

1 answer:

aalyn [17]2 years ago

4 0

Answer:

T=1.566 N.m

Explanation:

Given that:

rotational speed of the scrank shaft, N = 2500 rpm

power produced by one cylinder, P = 410 W

We know, in case of rotational power:

$P=\frac{2\pi.N.T}{60}$

where: T= torque

Substituting the respective values in the above eq.

$410=\frac{2\pi\times2500\times T}{60}$

$T= \frac{410\times60}{2\pi\times2500}$

T=1.566 N.m is the torque applied by the each piston of the engine.

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<span>Well, since it's in the shape of a wheel and the person walks around the edge of it, they must have a centripetal acceleration. Since a=v^2/r you can solve for "v" using 2.20 as your "a" and 59.5 as your "r" (r=half of the diameter).
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3 years ago

When the displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position, what fraction of the

KonstantinChe [14]

Answer:

The ratio is KE : TM = 0.75

Explanation:

from the question we are told that

The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position

Generally the total mechanical energy of the mass is mathematically represented as

$TM = \frac{1}{2} * k * A^2$

Here k is the spring constant , A is the total displacement of the the mass from maximum compression to maximum extension of the spring

Generally this total mechanical energy is mathematically represented as

$TM = KE + PE$

=> $KE = TM - PE$

Here the potential energy of the mass is mathematically represented as

$PE = \frac{1}{ 2} * k * [ x ]^2$

Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is

$x = \frac{A}{2}$

So

$PE = \frac{1}{ 2} * k * [ \frac{A}{2} ]^2$

So

$KE = \frac{1}{2} * k * A^2 - \frac{1}{2} * k * [\frac{A}{2} ]^2$

=> $KE = \frac{1}{2} * k * A^2 - \frac{1}{8} * k * A ^2$

=> $KE = 0.375 * k * A^2$

So the ratio of $KE : TM$ is mathematically represented as

$\frac{KE}{TM} = \frac{0.375 k A^2 }{0.5 k A^2}$

=> $\frac{KE}{TM} = 0.75$

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