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sineoko [7]
2 years ago
7

Two particles, one with charge -5.45 × 10^-6 C and one with charge 4.39 × 10^-6 C, are 0.0209 meters apart. What is the magnitud

e of the force that one particle exerts on the other? Two new particles, which have identical positive charge q3, are placed the same 0.0209 meters apart, and the force between them is measured to be the same as that between the original particles. What is q3?
Physics
1 answer:
Levart [38]2 years ago
3 0

Explanation:

Given that,

Charge 1, q_1=-5.45\times 10^{-6}\ C

Charge 2, q_2=4.39\times 10^{-6}\ C

Distance between charges, r = 0.0209 m

1. The electric force is given by :

F=k\dfrac{q_1q_2}{r^2}

F=9\times 10^9\times \dfrac{-5.45\times 10^{-6}\times 4.39\times 10^{-6}}{(0.0209)^2}

F = -492.95 N

2. Distance between two identical charges, r=0.0209\ m

Electric force is given by :

F=\dfrac{kq_3^2}{r^2}

q_3=\sqrt{\dfrac{Fr^2}{k}}

q_3=\sqrt{\dfrac{492.95\times (0.0209)^2}{9\times 10^9}}

q_3=4.89\times 10^{-6}\ C

Hence, this is the required solution.

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Answer:

False

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Explanation:

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2 years ago
A 1,600 kg train car rolling freely on level track at 16 m/s bumps into a 1.0 × 103 kg train car moving at 10.0 m/s in the same
emmainna [20.7K]
This item is solved through the concept of the conservation of momentum which states that the momentum before and after collision should be equal. 
                                   momentum = mass x velocity
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3 years ago
The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 L of blood. Assume tha
Yuri [45]

Answer:

(a) 1.257 x 10^5 J

(b) 1.456 Watt

Explanation:

Volume of blood, v = 7500 L = 7.5 m^3

Height, h = 1.63 m

density of blood, d = 1.05 x 10^3 kg/m^3

(a) work done = m x g x h

W = v x d x g x h = 7.5 x 1.05 x 1000 x 9.8 x 1.63 = 1.257 x 10^5 J

(b) time = 1 day = 24 x 60 x 60 s = 86400 seconds

Power = Work / time = 1.257 x 10^5 / 86400 = 1.456 Watt

6 0
2 years ago
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Given :

Initial speed of car A is 15 m/s and initial speed of car B is zero.

Final speed of car A is zero and final speed of car B is 10 m/s.

To Find :

What fraction of the initial kinetic energy is lost in the collision.

Solution :

Initial kinetic energy is :

K.E_i = \dfrac{15^2m}{2} + 0\\\\K.E_i = \dfrac{225 m}{2}

Final kinetic energy is :

K.E_f = \dfrac{10^2m}{2} + 0\\\\K.E_f = \dfrac{100m}{2}

Now, fraction of initial kinetic energy loss is :

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Therefore, fraction of initial kinetic energy loss in the collision is 1.25 .

6 0
2 years ago
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