1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
erik [133]
2 years ago
14

The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.724 W/m^2. What is the maximum v

alue of the magnetic field in the wave?
a. 77.9 nT
b. 55.1 nT
c. 38.9 nT
d. 108 nT
Physics
2 answers:
alexira [117]2 years ago
8 0

Answer:

a. 77.9 nT

Explanation:

Given;

average value of the electromagnetic wave, E = 0.724 W/m^2

The average value of a Poynting vector is given by;

S = \frac{1}{2\mu_o} *E_oB_o\\\\But, E_o = cB_o\\\\S = \frac{1}{2\mu_o} *cB_o^2\\\\cB_o ^2= 2\mu_o S\\\\B_o^2 = \frac{2\mu_o S}{c}\\\\B_o = \sqrt{ \frac{2\mu_o S}{c}} \\\\B_o =   \sqrt{ \frac{2(4\pi*10^{-7}) (0.724)}{3*10^8}}\\\\B_o = 7.79*10^{-8} \ T\\\\B_o = 77.9*10^{-9} \ T\\\\B_o = 77.9 \ nT

Therefore, the maximum value of the magnetic field in the wave is 77.9 nT.

Gala2k [10]2 years ago
5 0

Answer:

A I think : )

Explanation:

You might be interested in
Which statement is true about an object that sinks in water?
Rina8888 [55]
A > Is the correct answer :)
5 0
3 years ago
Read 2 more answers
Certain neutron stars (extremely dense stars) are believed to be rotating at about 1000 rev/s. If such a star has a radius of 14
Sever21 [200]

Answer:

minimum mass of the neutron star = 1.624 × 10^30 kg

Explanation:

For  a material to remain on the surface of a rapidly rotating neuron star, the magnitude oĺf the gravitational acceleration on the material must be equal to the magnitude of the centripetal acceleration of the rotating neuron star.

This can be represented by the explanations in the attached document.

minimum mass of the neutron star = 1.624 × 10^30 kg

8 0
3 years ago
A ball is rolled uphill a distance of 12 meters before it slows, stops, and begins to roll back. The ball rolls downhill 20 mete
Sedaia [141]
The ball rolled a distance of
d = 12m + 20m.
But the change of position is
x = + 12m - 20m
5 0
3 years ago
A spring stretches by 0.0208 m when a 3.39-kg object is suspended from its end. How much mass should be attached to this spring
sashaice [31]

Answer:

COMPLETE QUESTION

A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?

Explanation:

Given that,

Extension of spring

x = 0.0208m

Mass attached m = 3.39kg

Additional mass to have a frequency f

Let the additional mass be m

Using Hooke's law

F= kx

Where F = W = mg = 3.39 ×9.81

F = 33.26N

Then,

F = kx

k = F/x

k = 33.26/0.0208

k = 1598.84 N/m

The frequency is given as

f = ½π√k/m

Make m subject of formula

f² = ¼π² •(k/m

4π²f² = k/m

Then, m4π²f² = k

So, m = k/(4π²f²)

So, this is the general formula,

Then let use the frequency above

f = 3Hz

m = 1598.84/(4×π²×3²)

m = 4.5 kg

4 0
3 years ago
A ball rolls down the hill which has a vertical height of 15 m. Ignoring friction what would be the gravitational potential ener
trasher [3.6K]

a) Potential energy: 147 m [J]

The gravitational potential energy of an object is given by

U=mgh

where

m is its mass

g=9.8 m/s^2 is the acceleration of gravity

h is the height of the object above the ground

In this problem,

h = 15 m

We call 'm' the mass of the ball, since we don't know it

So, the potential energy of the ball at the top of the hill is

U=(m)(9.8)(15)=147 m (J)

b) Velocity of the ball at the bottom of the hill: 17.1 m/s

According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:

U=K=\frac{1}{2}mv^2

where

v is the final velocity of the ball

We know from part a) that

U = 147 m

Substituting into the equation above,

147 m = \frac{1}{2}mv^2

And re-arranging for v, we find the velocity:

v=\sqrt{2\cdot 147}=17.1 m/s

5 0
3 years ago
Other questions:
  • The rope of the length 6 m is fixed to a vertical wall in the horizontal x-direction. We send one transverse pulse on a rope in
    6·1 answer
  • Which of the following statement is true?
    14·1 answer
  • Two speakers create identical
    6·1 answer
  • How does volcanic ash in earth's atmosphere affect solar radiation?
    8·2 answers
  • According to newton's first law, an object traveling in a circle does not have a force acting on it.
    7·1 answer
  • The function H(t) = -16t^2 + vt + s shows the height H (t), in feet, of a projectile launched vertically from s feet above the g
    15·1 answer
  • Plz help!! WILL MARK BRAINLIEST :)
    15·1 answer
  • How do I determine the amount of time taken to empty a large tank full of water?
    9·1 answer
  • A sled is slowing down at the bottom of a snowy hill.
    13·1 answer
  • Which quantity must be the same for two bodies in thermal equilibrium?
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!