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Firdavs [7]
3 years ago
10

A bullet is fired horizontally with an initial velocity of 144.7 m/s from a tower 11 m high. if air resistance is negligible, wh

at is the horizontal distance the bullet travels before hitting the ground?
Physics
1 answer:
scoundrel [369]3 years ago
6 0
H = 11 m, the vertical distance that the bullet falls.
Initial vertical velocity  = 0
Horizontal velocity  = 144.7 m/s

The time, t, taken to fall 11 m is given by
(1/2)*(9.8 m/s²)*(t s)² = (11 m)
4.9t² = 11
t = 1.4983 s

If aerodynamic resistance is ignored, the horizontal distance traveled before the bullet hits the ground is
d = (144.7 m/s)*(1.4983 s) = 216.8 m

Answer: 216.8 m
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given,

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Answer:

\large \boxed{42\, \mu \text{C}}$

Explanation:

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The charges are identical, so we can write the formula as

F=k\dfrac{q^{2}}{r^2}

\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}

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3 years ago
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Answer:

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Explanation:

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