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Naddik [55]
2 years ago
8

Diffuse reflection occurs when the size of surface irregularities is

Physics
1 answer:
erik [133]2 years ago
7 0

Answer:

Varies from point to point on the surface of the material.

Explanation:

Diffusion reflection is the reflection of light due to irregular reflection angles on the material surface. It might be as a result of microscopic irregularities on the material surface or just plain non reflecting ability of the material.

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The conservation of momentum is most closely related to
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<h2>B) Newton's 2nd law</h2>

Explanation:

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Calculate the momentum of a 1,500 kg car traveling at 6 m/s.
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Answer: 9000 kgm/s

Explanation:

Mass of car = 1500 kg

Speed by which car moves = 6 m/s. Momentum of the car = ?

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= 1500kg x 6m/s

= 9000 kgm/s

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3 years ago
Ag,Au and Cu are called coinage metals why plzzzz hurry its urgent plzzz​
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3 years ago
A solenoid 25.0 cmcm long and with a cross-sectional area of 0.550 cm^2 contains 460 turns of wire and carries a current of 90.0
ankoles [38]

Answer:

a.  B = 0.20T

b.  u = 17230.6 J/m³

c.  E = 0.236J

d.  L = 5.84*10^-5 H

Explanation:

a. In order to calculate the magnetic field in the solenoid you use the following formula:

B=\frac{\mu_o n i}{L}               (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

n: turns of the solenoid = 460

L: length of the solenoid = 25.0cm = 0.25m

i: current  = 90.0A

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

B=\frac{(4\pi*10^{-7}T/A)(460)(90.0A)}{0.25m}=0.20T

The magnetic field in the solenoid is 0.20T

b. The magnetic permeability of air is approximately equal to the magnetic permeability of vacuum. To calculate the energy density in the solenoid you use:

u=\frac{B^2}{2\mu_o}=\frac{(0.20T)^2}{2(4\pi*10^{-7}T/A)}=17230.6\frac{J}{m^3}

The energy density is 17230.6 J/m³

c. The total energy contained in the solenoid is:

E=uV           (2)

V is the volume of the solenoid and is calculated by assuming the solenoid as a perfect cylinder:

V=AL

A: cross-sectional area of the solenoid = 0.550 cm^2 = 5.5*10^-5m^2

V=(5.5*10^{-5}m^2)(0.25m)=1.375*10^{-5}m^3

Then, the energy contained in the solenoid is:

E=(17230.6J/m^3)(1.375*10^{-5}m^3)=0.236J

The energy contained is 0.236J

d. The inductance of the solenoid is calculated as follow:

L=\frac{\mu_o N^2 A}{L}=\frac{(4\pi*10^{-7}T/A)(460)^2(5.5*10^{-5}m^2)}{0.25m}\\\\L=5.84*10^{-5}H

The inductance of the solenoid is 5.84*10^-5 H

3 0
3 years ago
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