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3 years ago

7

In a new scenario, the block only makes it (exactly) half-way through the rough spot. How far was the spring compressed from its

unstretched length

1 answer:

Mila [183]3 years ago

4 0

Answer: hello below is the missing part of your question

A mass m = 10 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5029 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μk = 0.49. The mass leaves the spring at a speed v = 3.4 m/s.

answer

x = 0.0962 m

Explanation:

<em>First step :</em>

Determine the length of the rough patch/spot

F = Uₓ (mg)

and w = F.d = Uₓ (mg) * d

hence;

d( length of rough patch) = w / Uₓ (mg) = 46.55 / (0.49 * 10 * 9.8) = 0.9694 m

<em>next : </em>

work done on unstretched spring length

Given that block travels halfway i.e. d = 0.9694 / 2 = 0.4847 m

w' = Uₓ (mg) * d

= 0.49 * 10 * 9.81 * 0.4847 = 23.27 J

also given that the Elastic energy of spring = work done ( w')

1/2 * kx^2 = 23.27 J

x = $\sqrt{\frac{2*23.27}{5029} }$ = 0.0962 m

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Answer:

The time it takes the ball to stop is 0.021 s.

Explanation:

Given;

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applied force, F = 520 N

Apply Newton's second law of motion, to determine the time it takes the ball to stop;

$F = ma = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{0.72 \ \times \ 15}{520} \\\\t = 0.021 \ s \\$

Therefore, the time it takes the ball to stop is 0.021 s.

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3 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

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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 $i_2$

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

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$q_0=0.01166\ c$

Finally

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

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3 years ago

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3 years ago