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rewona [7]
3 years ago
7

Calculate the magnitude of the force exerted by each wire on a 1.20-m length of the other.

Physics
1 answer:
Zielflug [23.3K]3 years ago
4 0

Incomplete question.The complete question is attached below as screenshot along with figure

Answer:

F=6.00*10^{-6}N

Force is repulsive

Explanation:

Given data

Current I₁=5.00A

Current I₂=2.00A

Length L=1.20 m

Radius r=0.400m

To find

Force F

Solution

As the force is repulsive because currents are in opposite direction

From repulsive force we know that:

F=\frac{u_{o}I_{1}I_{2}L}{2\pi r}

Substitute the given values

F=\frac{u_{o}(5.00A)(2.00A)(1.20m)}{2\pi (0.400m)}\\ F=6.00*10^{-6}N

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krek1111 [17]

Answer:

Explanation: In DC circuit, the current will flow for a short time, which is required to charge the capacitor. Once you switch it on, it spikes and the gradually decreases to almost zero (0) as the capacitor becomes fully charged.

In an AC circuit, the circuit acts as if the current is flowing throw the plates whereas is not actually flowing. The circuit acts like the AC is flowing through the capacitor.

8 0
3 years ago
An object is dropped from a height of 25 meters. At what velocity will it hit the ground?
Goryan [66]
You can use Vf^2-Vi^2 = 2ax

Vf^2 - 0 = 2(9.81)(25)

Or you can use energy

mgh = 1/2mv^2

2gh =v^2

Same thing
6 0
3 years ago
Part 1: Fill in the SI unit for each of the following measurements. 1. Time: 2. Length: 3. Mass: 4. Temperature:
Alik [6]

Answer:

1. Time: Second

2. Length: Meter

3. Mass: Kilogram

4. Temperatur: Kelvin

Explanation:

...

3 0
3 years ago
A box is placed on a 30o frictionless incline. What is the acceleration of the box as it slides down the incline
balandron [24]

Answer:

<em>2.78m/s²</em>

Explanation:

Complete question:

<em>A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?</em>

According to Newton's second law of motion:

\sum F_x = ma_x\\F_m - F_f = ma_x\\mgsin\theta - \mu mg cos\theta = ma_x\\gsin\theta - \mu g cos\theta = a_x\\

Where:

\mu is the coefficient of friction

g is the acceleration due to gravity

Fm is the moving force acting on the body

Ff is the frictional force

m is the mass of the box

a is the acceleration'

Given

\theta = 30^0\\\mu = 0.25\\g = 9.8m/s^2

Required

acceleration of the box

Substitute the given parameters into the resulting expression above:

Recall that:

gsin\theta - \mu g cos\theta = a_x\\

9.8sin30 - 0.25(9.8)cos30 = ax

9.8(0.5) - 0.25(9.8)(0.866) = ax

4.9 - 2.1217 = ax

ax = 2.78m/s²

<em>Hence the acceleration of the box as it slides down the incline is 2.78m/s²</em>

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3 years ago
Found throughout the nervous system, __________ aid and support the function of neurons.
Nonamiya [84]

Answer:

Glia

Explanation:

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2 years ago
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