All categories

3 years ago

If 2.0 x 10^-4 C charge passes a point in 5.0 x 10^-5 s, what is the rate of current flow?

1 answer:

3 0

Current is defined as the rate of charge flowing a point every second. Having a current of 1 Ampere signifies 1 Coulomb is flowing in a circuit every second. It is measured by the use of an ammeter which is positioned in series to the component to be measured. The current in the problem is calculated as follows:

I = 2.0 x 10^-4 C / 5.0 x 10^-5 s
<span>I = 4 A or 4.0 x 10^0 A</span>

You might be interested in

What will a spring scale read for the weight of a 57.0-kg woman in an elevator that moves upward with constant speed of 5.0 m/s

Tanzania [10]

The spring scale will read 559 Newton's or 125.7 pounds.

4 0

3 years ago

Longitudinal waves transfer energy ___________ to the direction of the wave motion.

viva [34]

Longitudinal waves transfer energy parallel to the direction of the wave motion

4 0

3 years ago

Read 2 more answers

Why does the amount of damage to a car depend on speed?

Zina [86]

if you are going slow, there wont be much of an effect or not any damage.

3 0

3 years ago

A helicopter has blades of length 4.0 m rotating at 3.0 rev/s in a horizontal plane.If the vertical component of the Earth’s mag

VARVARA [1.3K]

Answer:

Induced emf, $\epsilon=9.79\times 10^7\ volts$

Explanation:

Given that,

Length of the helicopter, l = 4 m

Angular speed of the helicopter, $\omega=3\ rev/s=18.84\ rad/s$

The vertical component of the Earth’s magnetic field is, $B=6.5\times 10^5\ T$

We need to find the induced emf between the tip of a blade and the hub. The induced emf in terms of angular velocity of an rotating object is given by :

$\epsilon=\dfrac{1}{2}B\omega l^2$

$\epsilon=\dfrac{1}{2}\times 6.5\times 10^5\times 18.84\times (4)^2$

$\epsilon=9.79\times 10^7\ volts$

So, the induced emf between the tip of a blade and the hub is $\epsilon=9.79\times 10^7\ volts$ . Hence, this is the required solution.

5 0

3 years ago

Coulomb measured the deflection of sphere A when spheres A and B had equal charges and were a distance d apart. He then made the

sdas [7]

Answer:

The new distance is d = 0.447 d₀

Explanation:

The electric out is given by Coulomb's Law

F = k q₁ q₂ / r²

This electric force is in balance with tension.

We reduce the charge of sphere B to 1/5 of its initial value ( $q_{B}$ =q₂ = q₂ / 5) than new distance (d = n d₀)

dat

q₁ = $q_{A}$

q₂ = $q_{B}$

r = d₀

In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points

F = k q₁ q₂ / d₀²

F = k q₁ (q₂ / 5) / (n d₀)²

.k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)

5 n² = 1

n = √ 1/5

n = 0.447

The new distance is

d = 0.447 d₀

6 0

3 years ago