HELP PLEASE!!

In the United States, household electric power is provided at a frequency of 60 HzHz, so electromagnetic radiation at that frequ

grigory [225]

Answer:

the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

Explanation:

Given the data in the question;

To determine the maximum intensity of an electromagnetic wave, we use the formula;

$I$ = $\frac{1}{2}$ ε₀cE $_{max$ ²

where ε₀ is permittivity of free space ( 8.85 × 10⁻¹² C²/N.m² )

c is the speed of light ( 3 × 10⁸ m/s )

E $_{max$ is the maximum magnitude of the electric field

first we calculate the maximum magnitude of the electric field ( E $_{max$ )

E $_{max$ = 350/f kV/m

given that frequency of 60 Hz, we substitute

E $_{max$ = 350/60 kV/m

E $_{max$ = 5.83333 kV/m

E $_{max$ = 5.83333 kV/m × ( $\frac{1000 V/m}{1 kV/m}$ )

E $_{max$ = 5833.33 N/C

so we substitute all our values into the formula for intensity of an electromagnetic wave;

$I$ = $\frac{1}{2}$ ε₀cE $_{max$ ²

$I$ = $\frac{1}{2}$ × ( 8.85 × 10⁻¹² C²/N.m² ) × ( 3 × 10⁸ m/s ) × ( 5833.33 N/C )²

$I$ = 45 × 10³ W/m²

$I$ = 45 × 10³ W/m² × ( $\frac{1 kW/m^2}{10^3W/m^2}$ )

$I$ = 45 kW/m²

Therefore, the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

7 0

3 years ago

As a car accelerates its tires increase their angular velocity from 114 rad/s to 131 rad/s in 0.43 seconds. What is the angular

e-lub [12.9K]

I think it's 39.53

(please do calculate it. I am not completely sure)

5 0

3 years ago

Read 2 more answers

The flux through the coils of a solenoid changes from 2.57.10-5 Wb to 9.44.10-5 Wb in 0.0154 s. If 4.08 V of EMF is generated, h

Vinil7 [7]

Hello!

We can use Faraday's Law of Electromagnetic Induction to solve.

$\epsilon = -N \frac{d\Phi_B}{dt}$

ε = Induced emf (4.08 V)
N = Number of loops (?)

$\Phi_B$ = Magnetic Flux (Wb)
t = time (s)

**Note: The negative sign can be disregarded for this situation. The sign simply shows how the induced emf OPPOSES the current.

Now, we know that $\frac{d\Phi_B}{dt}$ is analogous to the change in magnetic flux over change in time, or $\frac{\Delta \Phi_B}{\Delta t}$ , so:
$\epsilon = N \frac{\Delta \Phi_B}{\Delta t}\\\\\epsilon = N \frac{\Phi_{Bf} - \Phi_{Bi}}{\Delta t}$

Rearrange the equation to solve for 'N'.

$N = \frac{\epsilon}{ \frac{\Phi_{Bf} - \Phi_{Bi}}{\Delta t}}$

Plug in the given values to solve.

$N = \frac{4.08}{ \frac{9.44*10^{-5} - 2.57*10^{-5}}{0.0154}} = 914.585 = \boxed{915 \text{ coils}}$

**Rounding up because we cannot have a part of a loop.

4 0

2 years ago

An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference

kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

$K=qV$ (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

$K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J$

The kinetic energy of the particle is also:

$K=\frac{1}{2}mv^2$ (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}$

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

$|F_B|=qvBsin(\theta)$

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

$|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N$

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

$r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm$

the radius of the trajectory of the electron is 10.1 cm

6 0

4 years ago

which of the following are examples of circular motion? A. Roller skating down a hill. B. A race car going around a rounded curv

Katena32 [7]

Answer:

C. Earth going around the sun.

Explanation:

Circular motion should have a center to repeat its motion.

6 0

3 years ago