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2 years ago

8

After a car accident, the skid marks for one car measured 190 feet. Use the formula s = √ 24 d , where d represents the length o

f the skid mark (in feet) and s represents the speed of the car before the brakes were applied, to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

1 answer:

jarptica [38.1K]2 years ago

3 0

Answer:

s = 67.5 m/s

Explanation:

Skid mark for the car = 190 ft

given equation of speed and distnace

$s = \sqrt{24d}$

d is the distance

s is the speed

$s = \sqrt{24\times 190}$

$s = \sqrt{4560}$

$s =67.52 m/s$

rounding off the answer to the nearest tenth

s = 67.5 m/s

The speed of the car before it stop is equal to s = 67.5 m/s

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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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Answer:

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<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 $t, t_1, t_2$

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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:

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Squaring and adding the last two equations, and knowing that

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Operating

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Solving for $i_2$

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