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Lera25 [3.4K]
3 years ago
10

A point charge with a charge q1 = 4.00 μC is held stationary at the origin. A second point charge with a charge q2 = -4.10 μC mo

ves from the point x= 0.170 m , y=0 to the point x= 0.270 m , y= 0.270 m. How much work is done by the electric force on q2?
Physics
1 answer:
GrogVix [38]3 years ago
4 0

Answer:

W = -0.480 J

Explanation:

given,

q₁ = 4 μC

q₂ = -4.10 μC

W = kq_1q_2(\dfrac{1}{a}+\dfrac{1}{b})

b = \sqrt{(0.27-0)^2+(0.27-0)^2}

b = 0.381

k = 8.99 × 10⁹ Nm²/C²

W = 8.99\times 10^9\times 4\times 10^{-6}\times (-4.1 \times 10^{-6})(\dfrac{1}{0.17}+\dfrac{1}{0.381})

W = [-147.436\times (5.88-2.62)\times 10^{-3}]J

W = -0.480 J

Work done by the electric force W = -0.480 J

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The coefficient of linear expansion, given that the length of the pipe increased by 1.5 cm is 1.67×10¯⁵ /°F

<h3>How to determine the coefficient of linear expansion</h3>

From the question given above, the following data were obtained

  • Original diameter (L₁) = 10 m
  • Change in length (∆L) = 1.5 cm = 1.5 / 100 = 0.015 m
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The coefficient of linear expansion can be obtained as illustrated below:

α = ∆L / L₁∆T

α = 0.015 / (10 × 90)

α = 0.015 / 900

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3 0
1 year ago
A disk of mass M and radius R rotates at angular velocity ω0. Another disk of mass M and radius r is dropped on top of the rotat
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Answer:

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L_f = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega

now from above equation we have

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now we have

\omega = \frac{MR^2\omega_o}{(MR^2 + Mr^2)}

\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}

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The two particles are both moving to the right. Particle 1 catches up with particle 2 and collides with it. The particles stick
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Answer:

c. vf is greator than v2, but less than v1

Explanation:

The principle of conservation of linear momentum states that when two or more bodies act upon one another, their total momentum remains constant.

In a system of colliding bodies the total momentum of the system just before the collision is the same as the total momentum just after the collision.  

Collisions in which the kinetic energy is conserved are called elastic collision.

Collisions in which the kinetic energy is not conserved are called inelastic collisions.  If the two objects stick together after the collision and move with a common velocity, the collision is said to be perfectly inelastic.

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Therefore,

a. vf = v2 is wrong because vf is greater than v2

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c. vf is greater than v2, but less than v1 is correct.

d. vf = v1 is wrong because vf is less than v1

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