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hjlf
3 years ago
10

At some instant and location, the electric field associated with an electromagnetic wave in vacuum has the strength 71.9 V/m. Fi

nd the magnetic field strength B, the total energy density u, and the power flow per unit area, all at the same instant and location.
Physics
1 answer:
kirza4 [7]3 years ago
6 0

Answer:

a) Magnetic field strength, B = 2.397 * 10⁻⁷ T

b) Total energy density, U = 4.58 * 10⁻⁸ J/m³

c) Power flow per unit area, S = 13.71 W/m²

Explanation:

a) Electric field strength, E = 71.9 V/m

The relationship between the Electric field strength and the magnetic field strength in vacuum is:

E = Bc where c = 3.0 * 10⁸ m/s

71.9 = B * 3.0 * 10⁸

B = 71.9 / (3.0 * 10⁸)

B = 23.97 * 10⁻⁸

B = 2.397 * 10⁻⁷ T

b) Total Energy Density:

U = \frac{1}{2} \epsilon_0E^2 +  \frac{1}{2} \frac{B^2}{\mu_0} \\U = \frac{1}{2}* 8.85 * 10^{-12}*71.9^2 +  \frac{1}{2} \frac{(2.397*10^{-7})^2}{4\pi*10^{-7}}\\U = 2.29 * 10^{-8} + 2.29 * 10^{-8}\\U = 4.58 * 10^{-8} J/m^3

c)Power flow per unit area

S = \frac{1}{\mu_0} EB\\S = \frac{1}{4\pi * 10^{-7} } * 71.9 * 2.397 * 10^{-7}\\S = 13.71 W/m^2

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saveliy_v [14]

Answer:

d. The scale's resolution is too low to read the change in mass

Explanation:

If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:

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since,

w = Fd

dw = Fdx

integrating and using value of F, we get:

ΔE = (0.5)kx²

where,

ΔE = Energy added to spring

k = spring constant

x = displacement

The spring constant is typically in range of 4900 to 29400 N/m.

So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:

ΔE = (0.5)(29400 N/m)(1 m)²

ΔE = 1.47 x 10⁴ J

Now, if we convert this energy to mass from Einstein's equation, we get:

ΔE = Δmc²

Δm = ΔE/c²

Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²

<u>Δm =  4.9 x 10⁻¹³ kg</u>

As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

Thus, it can not determine such a small change. So, the correct option is:

<u>d. The scale's resolution is too low to read the change in mass</u>

8 0
3 years ago
Suppose you monitor a large number (many thousands) of stars over a period of 3 years, searching for planets through the transit
nasty-shy [4]

Answer:

2. You must be able to precisely measure variations in the star's brightness with time.

5. As seen from Earth, the planet's orbit must be seen nearly edge–on (in the plane of our line-of-sight).

6. You must repeatedly obtain spectra of the star that the planet orbits.

Explanation:

The transit method is a very important and effective tool for discovering new exoplanets (the planets orbiting other stars out of the solar system). In this method the stars are observed for a long duration. When the exoplanet will cross in front of theses stars as seen from Earth, the brightness of the star will dip. To observe this dip following conditions must be met:

1. The orbit of the planet should be co-planar with the plane of our line of sight. Then only its transition can be observed.

2. The brightness of the star must be observed precisely as the period of transit can be less than a second as seen from Earth. Also the dip in brightness depends on the size of the planet. If the planet is not that big the intensity dip will be very less.

3. The spectrum of the star needs to be studied and observe during the transit and normally to find out the details about the planets.

4. Also, the orbital period should be less than the period of observation for the transit to occur at least once.

4 0
3 years ago
A particular planet has a moment of inertia of 9.74 × 1037 kg ⋅ m2 and a mass of 5.98 × 1024 kg. Based on these values, what is
malfutka [58]

Answer:  A) 6.38(10)^{6} m

Explanation:

The equation for the moment of inertia I of a sphere is:

I=\frac{2}{5}mr^{2} (1)

Where:

I=9.74(10)^{37}kg m^{2} is the moment of inertia of the planet (assumed with the shape of a sphere)

m=5.98(10)^{24}kg is the mass of the planet

r is the radius of the planet

Isolating r from (1):

r=\sqrt{\frac{5I}{2m}} (2)

Solving:

r=\sqrt{\frac{5(9.74(10)^{37}kg m^{2})}{2(5.98(10)^{24}kg)}} (3)

Finally:

r=6381149.077m \approx 6.38(10)^{6} m

Therefore, the correct option is A.

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Answer:

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EleoNora [17]

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now the component of velocity which is parallel to the surface and perpendicular to the surface is given as

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7 0
4 years ago
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