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qwelly [4]
3 years ago
9

The spring of a spring balance is 6.0 in. long when there is no weight on the balance, and it is 8.4 in. long with 4.0 lb hung f

rom the balance. How much work is done in stretching it from 6.0 in. to a length of 10.1 in.?
Physics
1 answer:
Sergeeva-Olga [200]3 years ago
3 0

Answer:

W = 55.12 J

Explanation:

Given,

Natural length = 6 in

Force = 4 lb,  stretched length = 8.4 in

We know,

F = k x

k is spring constant

4 = k (8.4-6)

k = 1.67 lb/in

Work done to stretch the spring to 10.1 in.

W =k\int_{6}^{10.1} x

W = \dfrac{k}{2}[x^2]_6^{10.1}

W = \dfrac{1}{2}\times 1.67\times (10.1^2-6.0^2)

W = 55.12 J

Work done in stretching spring from 6 in to 10.1 in is equal to 55.12 J.

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a tank circuit contains a capacitor and an inductor that produce 30 of reactance at the resonant frequency. the inductor has a q
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The total circuit current at the resonant frequency is 0.61 amps

What is a LC Circuit?

  • A capacitor and an inductor, denoted by the letters "C" and "L," respectively, make up an LC circuit, also referred to as a tank circuit, a tuned circuit, or a resonant circuit.
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Q =15 = (wL)/R

wL = 30 ohms = Xl

R = 2 ohms

Zs = R + jXl = 2 +j30 ohms where Zs is the series LR impedance

| Zs | = 30.07 <86.2° ohms

Xc = 1/(wC) = 30 ohms

The impedance of the LC circuit is found from:

Zp = (Zs)(-jXc)/( Zs -jXc)

Zp = (2+j30)(-j30)/(2 + j30-j30) = (900 -j60)2 = 450 -j30 = 451 < -3.81°

I capacitor = 277/-j30 = j9.23 amps

I Zs = 277/(2 +j30) = (554 - j8,310)/904 = 0.61 - j9.19 amps

I net = I cap + I Zs = 0.61 + j0.04 amps = 0.61 < 3.75° amps

Hence, the total circuit current at the resonant frequency is 0.61 amps

To learn more about LC Circuit from the given link

brainly.com/question/29383434

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