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Zepler [3.9K]
4 years ago
15

Tom has built a large slingshot, but it is not working quite right. He thinks he can model the slingshot like an ideal spring wi

th a spring constant of 35.0 N/m.35.0 N/m. When he pulls the slingshot back 0.375 m0.375 m from a nonstretched position, it just does not launch its payload as far as he wants. His physics professor "helps" by telling him to aim for an elastic potential energy of 22.0 J.22.0 J. Tom decides he just needs elastic bands with a higher spring constant. By what factor does Tom need to increase the spring constant to hit his potential energy goal?
Physics
1 answer:
Amiraneli [1.4K]4 years ago
7 0

Answer:

8.9

Explanation:

We can start by calculating the initial elastic potential energy of the spring. This is given by:

U=\frac{1}{2}kx^2 (1)

where

k = 35.0 N/m is the initial spring constant

x = 0.375 m is the compression of the spring

Solving the equation,

U=\frac{1}{2}(35.0)(0.375)^2=2.5 J

Later, the professor told the student that he needs an elastic potential energy of

U' = 22.0 J

to achieve his goal. Assuming that the compression of the spring will remain the same, this means that we can calculate the new spring constant that is needed to achieve this energy, by solving eq.(1) for k:

k'=\frac{2U'}{x^2}=\frac{2(22.0)}{0.375^2}=313 N/m

Therefore, Tom needs to increase the spring constant by a factor:

\frac{k'}{k}=\frac{313}{35}=8.9

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2 years ago
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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Answer:

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

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3 years ago
A rock is dropped off a cliff and falls the first half of the distance to the ground in 2.0 seconds. how long will it take to fa
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Answer:

The answer is 0.83 seconds.

Explanation:

The formula of free fall is following:

h=1/2*g*t^2

Where g=9.8 m/s^2 and t=2 seconds, the rock takes:

h=1/2*9.8*2^2=19.6

19.6 meters. This is the half distance of the cliff. The whole distance is 39.2 meters. So it takes:

39.2=1/2*9.8*t^2\\t^2=8\\t=2.83

2.83 second to fall down completely. The rock takes the second half of the cliff in 0.83 seconds

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Answer:

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Explanation:

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The wavelength of the second harmonic is calculated as follows;

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where;

L is the length of the pipe in the second harmonic

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L = \frac{\lambda}{4} + \frac{\lambda}{2} + \frac{\lambda}{4} \\\\L = \lambda

The frequency is calculated as follows;

F_1 = \frac{V}{\lambda} = \frac{340}{0.8} = 425 \ Hz

Therefore, the frequency of the second harmonic of the pipe is 425 Hz.

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