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3 years ago

A coin placed 30.8 cm from the center of a rotating, horizontal turntable slips when its speed is 50.8 cm/s.

Engineering

1 answer:

Anestetic [448]3 years ago

6 0

Answer:

a)(A) static friction

b) $\mu _s=0.085$

Explanation:

Given that

r = 30.8 cm

V = 50.8 cm/s

(A) static friction

At the condition of motion

Static friction = Centrifugal force

$\mu _s\ m\ g=\dfrac{mV^2}{r}$

$\mu _s\ g=\dfrac{V^2}{r}$

$\mu _s\times 980=\dfrac{50.8^2}{30.8}$

$\mu _s=0.085$

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Both A and B technicians are correct because both might be used to test fuses, according to technician B.

<h3>What is continuity?</h3>

The behavior of a function at a certain point or section is described by continuity. The limit can be used to determine continuity.

From the question:

We can conclude:

The technician claims that you may check for continuity using both an ohmmeter and a self-powered test light. Both might be used to test fuses, according to technician B.

Thus, both A and B technicians are correct because both might be used to test fuses, according to technician B.

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Many companies are combining their online business activities with an existing physical presence. In about 100 words, explain wh

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3 years ago

For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calcu

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Answer:

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

B = -242.5 cm^3 mol^-1

C = 25200 cm^6 mo1^-2

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b= $v_1$

from virial equation

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And

$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a= $v_2\\$

b= $v_1$

Now calculate V1 and V2 at given condition

$\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}$

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Solve for V1 by iterative or alternative cubic equation solver we get

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Similarly solve for state 2 at P2 = 50 bar we get

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$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

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After performing integration we get work done on the system is

W=12.62 kJ/mol

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dV=RT(-1/p^2+0+C')dP

Hence work done on the system is

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a= $v_2\\$

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W=12.59 kJ/mol

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Answer:

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