Three sticks are arranged in such a way that they form a right triangle. The

An electron with a charge value of 1.6 x 10-19 C is moving in the presence of an electric field of 400 N/C. What force does the

Yuliya22 [3333]

The force experienced by the electron is 4.0×10⁻¹⁷ N.

4 0

2 months ago

Jo, Daniel and Helen are pulling a metal ring. Jo pulls with a force of 100N in one direction and Daniel with a force of 140N in

inna [3103]

Answer:

She exerts a force of 40 N.

Explanation:

The fact that the ring remains stationary indicates that the forces are in equilibrium.

Let’s denote Jo's force as x.

The equation to consider is

140 = x + 100

x = 40

5 0

2 months ago

If an instalment plan quotes a monthly interest rate of 4%, the effective annual/yearly interest rate would be _____________. 4%

inna [3103]

Answer:

More than 48%

Explanation:

If the interest is calculated monthly based on the outstanding balance, it leads to an effective annual rate of...

(1 + 4%)^12 - 1 = 60.1%... more than 48%

4 0

3 months ago

A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum e

kicyunya [3294]

Answer:

Maximum emf = 5.32 V

Explanation:

Provided data includes:

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions each second

Magnetic field, B = 0.5 T

We are tasked to determine the maximum emf produced in the loop, which is founded on Faraday's law. The induced emf can be calculated by:

$\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t$

For the maximum emf, $\sin\omega t=1$

Therefore,

$\epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V$

Hence, the maximum emf generated in the loop is 5.32 V.

3 0

3 months ago