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defon
2 years ago
12

What would be the answer for this and how?

Physics
1 answer:
Veronika [31]2 years ago
7 0

Answer:

B. 6 cm

Explanation:

First, we calculate the spring constant of a single spring:

k = \frac{F}{\Delta x}\\

where,

k = spring constant of single spring = ?

F = Force Applied = 10 N

Δx = extension = 4 cm = 0.04 m

Therefore,

k = \frac{10\ N}{0.04\ m}\\k =  250\ N/m\\

Now, the equivalent resistance of two springs connected in parallel, as shown in the diagram, will be:

k_{eq} = k + k\\k_{eq} = 2k = 2(250\ N/m)\\k_{eq} = 500\ N/m\\

For a load of 30 N, applying Hooke's Law:

\Delta x = \frac{F}{k_{eq}}\\\\\Delta x = \frac{30\ N}{500\ N/m}\\\\\Delta x = 0.06\ m = 6\ cm\\

Hence, the correct option is:

<u>B. 6 cm</u>

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<h3>a. </h3>

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2.7 \frac{m}{s^2} = \omega ^2 * 110  \ m

\omega ^2 = \frac{2.7 \frac{m}{s^2}} {110 \ m}

\omega  = \sqrt{ 0.02454 \frac{rad^2}{s^2} }

\omega  = 0.1567 \frac{rad}{s}

Knowing that there are 2\pi \ rad in a revolution and 60 seconds in a minute.

\omega  = 0.1567 \frac{rad}{s}  \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}

\omega  = 1.496 \frac{rev}{min}

<h3>b. </h3>

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\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}

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Knowing that there are 2\pi \ rad in a revolution and 60 seconds in a minute.

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In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu
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i_2=3.61\ A

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

q, q_1, q_2 = charge of the capacitor in any time t, t_1, t_2

q_o = initial charge of the capacitor

\omega=angular frequency of the circuit

i, i_1, i_2 = current through the circuit in any time t, t_1, t_2

The charge in an LC circuit is given by

q(t) = q_0 \, cos (\omega t )

The current is the derivative of the charge

\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).

We are given

q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}

It means that

q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]

i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]

From eq 1:

\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}

From eq 2:

\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}

Squaring and adding the last two equations, and knowing that

sin^2x+cos^2x=1

\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1

Operating

\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2

Solving for q_o

\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}

Now we know the value of q_0, we repeat the procedure of eq 1 and eq 2, but now at the second time t_2, and solve for i_2

\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2

Solving for i_2

\displaystyle i_2=w\sqrt{q_o^2-q_2^2}

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

\displaystyle f=\frac{1}{4\pi}\ KHz

w=2\pi f=500\ rad/s

\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}

q_0=0.01166\ c

Finally

\displaystyle i_2=500\sqrt{0.01166^2-.006^2}

i_2=5\ A

3 0
3 years ago
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