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AysviL [449]
3 years ago
12

"What is the energy density (energy per cubic meter) carried by the magnetic field vector in a small region of space in a EM wav

e at an instant of time when the electric vector is a maximum of 3500V/m
Physics
1 answer:
Pachacha [2.7K]3 years ago
6 0

Answer:

The energy density is  Z  = 5.4 2 *10^{-5 } \  J/m^3

Explanation:

From the question we are told that

    The electric vector is  E =  3500 \ V/m

Generally the energy vector is mathematically represented as

      Z  =  0.5  *  \epsilon_o  *  E^2

Where  \epsilon_o is the permitivity of free space with the value  \epsilon_o  =  8.85*10^{-12} \  \  m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

substituting values

      Z  =  0.5  * 8.85 *10^{-12} *  3500^2

      Z  = 5.4 2 *10^{-5 } \  J/m^3

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The velocity of sound on a particular day outside is 331 meters/second. What is the frequency of a tone if it has a wavelength o
Levart [38]
I think c hope it help

3 0
3 years ago
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At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial
elena55 [62]

Answer : The partial pressure of SO_3 is, 67.009 atm

Solution :  Given,

Partial pressure of SO_2 at equilibrium = 30.6 atm

Partial pressure of O_2 at equilibrium = 13.9 atm

Equilibrium constant = K_p=0.345

The given balanced equilibrium reaction is,

2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)

The expression of K_p will be,

K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times (p_{O_2})}

Now put all the values of partial pressure, we get

0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}

p_{SO_3}=67.009atm

Therefore, the partial pressure of SO_3 is, 67.009 atm

6 0
3 years ago
How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n
grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

dSin\theta = m\lambda

Where

m= Any integer which represent the number of repetition of spectrum

\lambda= Wavelength

d = Distance between the slits.

\theta= Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

d = \frac{m\lambda}{sin\theta }

d = \frac{2*536*10^{-9}}{sin(24)}

d = 2.63*10^{-6}m

Therefore the number of lines per millimeter would be given as

\frac{1}{d} = \frac{1}{2.63*10^{-6} }

\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})

\frac{1}{d} = 379.4 lines/mm

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

6 0
3 years ago
A car is driving at 99 km/h, calculate the distance it travels in 70 minutes.
Lapatulllka [165]

Answer:

The distance the car travels is 115500 m in S.I units

Explanation:

Distance d = vt where v = speed of the car and t = time taken to travel

Now v = 99 km/h. We now convert it to S.I units. So

v = 99 km/h = 99 × 1000 m/(1 × 3600 s)

v = 99000 m/3600 s

v = 27.5 m/s

The speed of the car is 27.5 m/s in S.I units

We now convert the time t = 70 minutes to seconds by multiplying it by 60.

So, t = 70 min = 70 × 60 s = 4200 s

The time taken to travel is 4200 s in S.I units

Now the distance, d = vt

d = 27.5 m/s × 4200 s

d = 115500 m

So, the distance the car travels is 115500 m in S.I units

8 0
3 years ago
Explain why nuclear fission and nuclear fusion release large amounts of energy
levacccp [35]

Answer:

Because of the formula E=mc^2

Explanation:

In this problem we are describing two different processes:

  • Nuclear fission occurs when a heavy, unstable nucleus breaks apart into two or more lighter nuclei
  • Nuclear fusion occurs when two (or more) light nuclei fuse together producing a heavier nucleus

In both cases, the total mass of the final products is smaller than the total mass of the initial nuclei.

According to Einsten's formula, this mass difference has been converted into energy, as follows:

E=\Delta mc^2

where:

E is the energy released in the reaction

\Delta m is the mass defect, the difference between the final total mass and the initial total mass

c=3.0 \cdot 10^8 m/s is the speed of light

From the formula, we see that the factor c^2 is a very large number, therefore even if the mass defect \Delta m is very small, nuclear fusion and nuclear fission release huge amounts of energy.

8 0
3 years ago
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