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2 years ago

What name is given to the rate of flow of electric charge?

Physics

1 answer:

RSB [31]2 years ago

3 0

Answer:

<h2>Electric charge</h2>

Explanation:

The rate of the flow of electric charge is known as electric current. <u>By convention, the direction of electric current is always the direction of net flow of positive charge.</u>

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A bowling ball has a mass of 7.2 kg and a weight of 70.6 N. It moves down the bowling alley at 1 m/s and strikes a pin with a fo

Inga [223]

The answer is 15.0N its explained in newtons third law hope this helps:)

4 0

3 years ago

Read 2 more answers

You are working in cooperation with the Public Health department to design an electrostatic trap for particles from auto emissio

Mademuasel [1]

Answer:

Explanation:

Given that:

Charge (q) on the particle = 3 × 10⁻⁸ C

mass (m) of the particle = 6 × 10⁻⁹ kg

at a distance x = 15 cm , the velocity in the plate = 900 m/s²

For the square plate, the surface charged density σ = -8 × 10⁻⁶ C/m²

To start with calculating the electric field as a result of the square plate; we use the formula;

$E = \dfrac{\sigma }{2 \varepsilon_o}$

$E = \dfrac{8 \times 10^{-6} }{2 \times 8.85 \times 10^{-12}}$

$E = 4.51977 \times 10^5 \ V/m$

On the square plate; The electric force F = Eq

$F = (4.51977 \times 10^5 \ V/m )(3\times 10^{-8} \ C)$

$F = 1.3559 \times 10^{-2} \ N$

The acceleration $a =\dfrac{ F}{m}{$

$a = \dfrac{1.3559\times 10^{-2} \ N}{6 \times 10^{-9} \ Kg}$

$a = 2.25988 \times 10^6 \ m/s^2$

For the particle, the velocity at distance x = 7 m can be calculated by using the formula:

$(\dfrac{1}{2}) mv^2 = \Delta Vq$

$v^2 = \dfrac{2 Eq}{dm}$

$v^2 = \dfrac{2 * 4.51977 \times 10^5 \times 3 \times 10^{-8} }{0.07 \times 6\times 10^{-9} }$

$v^2 = 64568142.86 \ m/s$

$v =\sqrt{ 64568142.86 \ m/s}$

$\mathbf{v = 8.035 \times 10^3 \ m/s}$

From the calculation, we realize that the charge acting between the particle and the plate is said to be "opposite".

Hence, the force is an attractive force.

Similarly, there is a gradual increase exhibited by the velocity of the particle.

Therefore, the particles get to the detector, but the detector failed to get detect due to the velocity which is greater than 1000 m/s.

6 0

3 years ago

A girl throws a pebble into a deep well at 4.0 m/s (downward). It hits the water in 2.0 s. How far below the ground is the water

alexandr402 [8]

Answer:

27.6 m
13.8 m/s

Explanation:

(b) The initial velocity is added to that due to acceleration by gravity. The velocity is increased linearly by gravity at the rate of 9.8 m/s². The average velocity of the pebble will be its velocity halfway through the 2-second time period.* That is, it will be ...

4 m/s + (9.8 m/s²)(2 s)/2 = 13.8 m/s . . . . average velocity

(a) The distance covered in 2 seconds at an average velocity of 13.8 m/s is ...

d = vt

d = (13.8 m/s)(2 s) = 27.6 m

The water is about 27.6 m below ground.

_____

* We have chosen to make use of the fact that the velocity curve is linear, so the average velocity is half the sum of initial and final velocities:

vAvg = (vInit + vFinal)/2 = (vInit + (vInit +at))/2 = vInit +at/2

If you work this in a straightforward way, you would find distance as the integral of velocity, then find average velocity from the distance and time.

$\displaystyle d=\int_0^t{(v_0+at)}\,dt=v_0t+\dfrac{1}{2}at^2=t\left(v_0+a\dfrac{t}{2}\right)\\\\v_{avg}=\dfrac{d}{t}=v_0+a\dfrac{t}{2}\qquad\text{the formula we started with}$