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tatuchka [14]
3 years ago
13

Charge X has twice as much charge as particle Y. The two charges are placed near each other. Compared to the force on particle X

, what is the force on particle Y? Question 8 options: Four times less Two times More The same None of these
Physics
1 answer:
Art [367]3 years ago
6 0

The force on charge Y is the same as the force on charge X

Explanation:

We can answer this problem by applying Newton's third law of motion, which states that:

"When an object A exerts a force on object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"

In this problem, we can identify object A as charge X and object B as charge Y. The magnitude of the electrostatic force between them is given by

F=k\frac{q_x q_y}{r^2}  (1)

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_x, q_y are the two charges

r is the separation between the two charges

According to Newton's third law, therefore, the magnitude of the force exerted by charge X on charge Y is the  same as the force exerted by charge Y on charge X (and it is given by eq.(1)), however their directions are opposite.

Learn more about Newton's third law:

brainly.com/question/11411375

#LearnwithBrainly

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The mass would be the same

47kg on the moon as well

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During a softball game, a shortstop catches a ground ball. The action force is the ball pushing on the glove. What is the reacti
Tanzania [10]

Its the floor pushing up on the player, this one is correct

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The acceleration due to gravity on the moon is 1.6 m/s2, about a sixth that of Earth’s. Which accurately describes the weight of
tatyana61 [14]

The answer is C.  

An object on the moon is six times lighter than on Earth.


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A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.400 rev
saw5 [17]

Answer:

The final angular velocity is rev/s is 0.293 rev/s.

Explanation:

Given;

mass of the merry-go-round, m₁ = 120 kg

radius of the merry-go-round, r = 1.8 m

initial angular velocity, ω = 0.4 rev/s

mass of the child, m₂ = 22 kg

Apply the principle of conservation angular momentum to determine the final angular velocity;

I_i= I_f\\\\\frac{1}{2} m_1r^2 \omega _i = \frac{1}{2} m_1r^2 \omega _f + m_2r^2 \omega _f\\\\ \frac{1}{2} m_1r^2 \omega _i =( \frac{1}{2} m_1r^2  + m_2r^2 )\omega _f\\\\\omega _f = \frac{ \frac{1}{2} m_1r^2 \omega _i}{\frac{1}{2} m_1r^2  + m_2r^2} \\\\\omega _f = \frac{ \frac{1}{2} m_1 \omega _i}{\frac{1}{2} m_1  + m_2}\\\\\omega _f = \frac{0.5 \ \times \ 120\ kg \ \times \ 0.4\ rev/s}{0.5 \ \times 120\ kg \ \ + \ \ 22 \ kg} \\\\\omega _f = 0.293 \ rev/s\\

Therefore, the final angular velocity is rev/s is 0.293 rev/s.

3 0
3 years ago
An ideal gas occupies a volume of 0.60 m^3 at 5.0 ATM and 400. K. What volume does it occupy at 4.0 ATM and a temperature of 200
sattari [20]

Answer:0.375m3

Explanation:This is general gas law. The expression is given by P1V1/T1 =P2V2/T2

we are looking for V2 so let's make it the change of subject of formula

V2= P1xV1xT2/P2T1

= 5 x 0.6 x 200/400 x 4

= 0.375m3

7 0
3 years ago
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