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3 years ago

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A bullet is fired horizontally from a 17.7 m high cliff at a speed of 482.0. What is the distance from the cliff that the bullet

will hit the ground below?
Please be sure to round to 1 decimal place. NUMBERS ONLY no units.

1 answer:

Marysya12 [62]3 years ago

5 0

Answer:

Horizontal distance=?m

Explanation:

Horizontal velocity,u=482ms⁻¹

Height of the cliff=17.7m

Horizontal distance,R=?

R=v×√2h/g

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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

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

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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 $q_o$

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

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Solving for $i_2$

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Finally

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