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koban [17]
3 years ago
11

A charge of 6.65 mC is placed at each corner of a square 0.500 m on a side.

Physics
1 answer:
mixer [17]3 years ago
7 0

Answer:

Explanation:

Given that:

length of side , a = 0.5 m

charge , q = 6.65 mC

length of diagonal , d = 0.5 * sqrt(2)

d = 0.707 m

F is the force due to adjacent particle ,

F1 is the force due to diagonal particle

Now , for the net charge on a particle

Fnet = 2 * F * cos(45) + F1

Fnet = 2*cos(45) * k * q^2/a^2 + k * q^2/d^2

Fnet = 9*10^9 * 0.00665^2 * (2* cos(45)/.5^2 + 1/.707^2)

Fnet = 3.05 *10^6 N

the magnitude of net force acting on each particle is 3.05 *10^6 N

part B)

for the direction of particle

d) along the line between the charge and the center of the square outward of the center

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3 years ago
A gas-filled balloon is submerged in a fluid. As a result, the balloon is subject to inward forces from all sides. How does the
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1 year ago
A​ hot-air balloon is 150 ft above the ground when a motorcycle​ (traveling in a straight line on a horizontal​ road) passes dir
Katyanochek1 [597]

Answer:

 dR/dt = 10.2 ft / s

Explanation:

Let's work this problem by finding the distance between the balloon and the motorcycle and then drift for the speed change of the distance

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      y = y₀ +v_{oy} t

Motorcycle

      x = v₀ₓ t

Distance, let's use Pythagoras' theorem

      R² = x² + y²

      R² = (v₀ₓ t)² + (y₀ + v_{oy} t)²

     v₀ₓ = 88 ft / s

     v_{oy} = 8 ft / s

     y₀ = 150 ft

     R² = (8 t)² + (150 + 8 t )²

     R² = 64 t² + (150 + 8t )²

This is the expression for the distance between the two bodies, the rate of change is the derivative with respect to time (d / dt)

         2RdR / dt = 64 2 t + 2 (150 + 8t) 8

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Let's calculate for the time of 10 s

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8 0
4 years ago
Please help fast!! I need this in less than 17 hours!
djyliett [7]

SOLUTION is given in attachment below.

7 0
3 years ago
You need to use your cell phone, which broadcasts an 830 MHz signal, but you’re in an alley between two massive, radio- wave-abs
Korvikt [17]

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