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3 years ago

5

Assume this piston has a maximum pressure rating of 2.5 pounds-force per square inch [psi]. The force applied to the cylindrical

piston is 100 newtons [N]. What is the diameter of the piston, in units of inches [in]

1 answer:

Leni [432]3 years ago

7 0

Explanation:

Always write what you know first about the problem

The surface of the piston that has the force applied is a circle

The area of a circle is pi x r^2

Pressure is Force / Area

Write this out

P = F / (pi x r^2)

solve for r

sqrt( F / (P x Pi)) = r

But dont forget that d = 2 x r

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Answer:

the workenergy principle is measured in joules

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A paper airplane is thrown westward at a rate of 6 m/s. The wind is blowing at 8 m/s towards the north. Which of the following d

Andrew [12]

since airplane is thrown towards west with speed 6 m/s

while air is blowing with speed 8 m/s towards north

so here the net speed of air plane will be the resultant of airplane speed and wind speed always

SO here we can say it would be a combination of vector along west with must be of length 6 m/s and other vector is towards north with is of length 8 m/s

so correct answer must be 1st option

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3 years ago

What is mean by the net displacement in transverse wave

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3 years ago

An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

5 0

3 years ago

Please help im stuck and can't find the answer sheet ANYWHERE :( PLEASEEEEEEEEEEE

Vladimir [108]

Answer:

All steps are 20 * 100 (break the rest into appropriate pieces)

You can multiply as follows

(2000) * ((3 * 60) + (2 * 60) + 60)

V = 2000 * 6 * 60) = 720,000 cm^3 = .72 m^3

.72 m^3 * 2400 kg / m^3 = 1728 kg

6 0

2 years ago