3) A ring of radius 15 cm is injected with an amount of charge of 3.5 × 10−12C, uniformly distributed along the ring. The ring rotates with an angular velocity of 17 rad/s. Find the power radiated by the system.

Explanation:

Given that

A ring of radius 15cm

r=0.15m

A charge of Q=3.5×10^-12C

Angular velocity is given as

w=17rad/s

Power is given as

P=IV, Electrical power

And the rate of charge I is given as

I=Q/t

Also, the potential difference V, for a point charge is given as

V=kQ/r

Therefore, power can be rewritten as

P=IV

P=(Q/t) × (kQ/r)

P=kQ²/tr

Note, t is the period and is given as

w=2πf

Then, f=1/t

w=2π/t,

So, 1/t=w/2π

P=kQ²/tr

P=(kQ²/r)(1/t) = (kQ²/r)(w/2π)

P=(kQ²/r)(w/2π)

K is a constant and it value is 9×10^9Nm²/C²

P=(9×10^9×3.5×10^-12×3.5×10^-12)(17/2π)

P=1.1025×10^-13×17/2π

P=2.98×10^-13 Watts

Given Information:

Radius = r = 15 cm = 0.15 m

Charge = q = 3.5x10⁻¹² C

Angular velocity = ω = 17 rad/s

Required Information:

Power radiated = P = ?

Power radiated = 1.987x10⁻¹² W

Explanation:

We know that power is given by

P =  VI

Where V is the potential due to charge and is given by

V = kq/r

Where k is the coulomb constant 8.98x10⁹ N.m²/C²

V = (8.98x10⁹*3.5x10⁻¹²)/0.15

V = 0.2095 V

we know that the rate of charge flow is current

I = q/t

where t is found by the relation

ω = 2π/t

t = 2π/ω

t = 2π/17

t = 0.369 s

so current is

I = q/t = 3.5x10⁻¹²/0.369

I = 9.485x10⁻¹² A

Therefore, the amount of power radiated is

P =VI

P = 0.2095*9.485x10⁻¹²

P = 1.987x10⁻¹² W

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong

the earths atmosphere is made up of nitrogen and oxygen. solids can turn into gasses through a process called sublimation. sublimation is the transition of a substance directly from the solid to the gas phase, without passing through the intermediate liquid phase.

hope this : d