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VLD [36.1K]
3 years ago
14

An electromagnetic wave with frequency 65.0hz travels in an insulating magnetic material that has dielectric constant 3.64 and r

elative permeability 5.18 at this frequency. the electric field has amplitude 7.20×10−3v/m.what is the intensity of the wave in a medium\?
Physics
1 answer:
torisob [31]3 years ago
4 0
Ans: Intensity = I = 5.8 * 10^{-8} W/m^2

Explanation:
First you need to find out the speed of Electromagnetic wave:

Since
v = \sqrt{ \frac{1}{\mu\epsilon} }


v = \sqrt{ \frac{1}{\mu_r\mu_ok\epsilon_o} }
\mu_r = 5.18 \\
\mu_o = 4 \pi * 10^{-7} \\
k = 3.64 \\
\epsilon_o = 8.85 * 10^{-12}

Plug in the values:
v = \sqrt{ \frac{1}{(5.18)(4\pi*10^{-7})(3.64)(8.85*10^{-12})} }
v = 6.9 * 10^7 m/s

Now that we have "v", we can use the following formula to find the intensity of wave:

I =  \frac{k\epsilon_o*v*E_{max}^2}{2}

I =  \frac{(3.64)(8.85*10^{-12})
*(6.9*10^7)*(7.20*10^{-3})^2}{2}

Intensity = I = 5.8 * 10^{-8} W/m^2

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1.989 × 10^30 kg and therefore the mass of the sun is 1.989 × 10^30 kg.
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3 years ago
Which positions experience low tide
MrMuchimi

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3 years ago
A rubber ball is dropped from a height of 8m. After strikingthe floor, the ball bounces to a height of 5m. a. If the ball had bo
kifflom [539]

Answer:

a) This means the collision between the ball and the floor is elastic.

b) This points to a perfectly inelastic collision between the ball and the floor as they stick together after collision

c) Check Explanation.

Explanation:

Collision of bodies are analysed according to whether both momentum and kinetic energy of the system is conserved, that is, if these two quantities before collision are equal to their values after collision.

In all types of collisions, momentum is usually conserved, but kinetic energy is conserved only in an elastic collision.

A ball dropped from a height of 8 m bounces up back to a height of 5 m.

a. If the ball had bounced to a height of 8m, how would you describe the collision between the ball and the floor?

The ball not bouncing back to a height of 8 m shows energy loss at some point in the total motion of the ball (most likely at the collision). If kinetic energy was conserved, the ball would bounce back up to the height at which it fell from (8 m) after the collision with the floor.

b. If the ball had not bounced at all, how would you describe the collision between the ball and the floor?

If the ball had not bounced at all, this means it lost all of its kinetic energy to the floor, and this points to a perfectly inelastic collision between the ball and the floor as they stick together after collision.

c. What happened to the energy lost by the ball during thecollision?

The energy lost during the collision is converted to another form, most likely responsible for some deformation on the ball & a minute deformation on the floor, converted to some form of heat as a result of the collision or into sound energy, usually, it's a combination of all This!

Hope this Helps!!!

5 0
4 years ago
A 3kg ball moving at 8 m/s strikes a 2kg ball at rest,if the collision is elastic,what is the speed of the lighter ball if the h
zloy xaker [14]

Answer:

Speed of lighter ball is 4 m/s.

Explanation:

Applying the principle of conservation of linear momentum,

momentum before collision = momentum after collision.

m_{1}u_{1} + m_{2} u_{2} = m_{1}v_{1} - m_{2}v_{2}

m_{1} = 3 kg, u_{1} = 8 m/s, m_{2} = 2 kg, u_{2} = 0 m/s ( since it is at rest), v_{1} = 2 m/s, v_{2} = ?

(3 x 8) + (2 x 0) = (8 x 2) - (2 x v_{2})

24 + 0 = 16 - 2v_{2}

2v_{2} = 16 - 24

2v_{2} = -8

v_{2} = \frac{-8}{2}

   = -4 m/s

This implies that the light ball moves at the speed of 4 m/s in the opposite direction of the heavier ball after collision.

7 0
3 years ago
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