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3 years ago

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An archer shot a 0.07 kg arrow at a target. The arrow accelerated at 4,000 m/s2 to reach a speed of 50.0 m/s as it left the bow.

During this acceleration, what was the net force on the arrow?
A) 280 N
B) .0000175 N
C) 57,142 N
D) 3.5 N

2 answers:

denpristay [2]3 years ago

3 0

Answer:

A. 280

Explanation:

saw5 [17]3 years ago

3 0

The answer is A ) 280 N

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A point charge q1=+5.00nC is at the fixed position x=0, y=0, z=0. You find that you must do 8.10×10−6J of work to bring a second

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The value of the second charge is 1.2 nC.

<h3>Electric potential</h3>

The work done in moving the charge from infinity to the given position is calculated as follows;

W = Eq₂

E = W/q₂

<h3>Magnitude of second charge</h3>

The magnitude of the second charge is determined by applying Coulomb's law.

$E = \frac{kq_2}{r^2} \\\\\frac{kq_2}{r^2} = \frac{W}{q_2} \\\\kq_2^2 = Wr^2\\\\q_2^2 = \frac{Wr^2}{k} \\\\q_2 = \sqrt{\frac{Wr^2}{k} } \\\\q_2 = \sqrt{\frac{(8.1 \times 10^{-6}) \times (0.04)^2}{9\times 10^9} } \\\\q_2 = 1.2 \times 10^{-9} \ C\\\\q_2 = 1.2 \ nC$

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Learn more about electric potential here: brainly.com/question/14306881

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2 years ago

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