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2 years ago

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A spaceship far from all other objects uses its impulse power system to attain a speed of 104 m/s. The crew then shuts off the p

ower. According to Newton's first law, what will happen to the motion of the spaceship from then on? (Select all that apply.)

1 answer:

asambeis [7]2 years ago

4 0

Answer:

Velocity remains the same at 104 m/s

Explanation:

According to Newton's 1st law of motion, an object subjected to no force or net force equal 0 would maintain its velocity. In our case the crew shuts off the power, spaceship is in space and far from all other objects (so no gravity whatsoever) would have no force acting on it. Therefore its velocity would stay the same at 104 m/s

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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

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Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

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