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Genrish500 [490]
3 years ago
15

What technology produced the results illustrated here, and what is this technology used for?

Physics
2 answers:
IrinaVladis [17]3 years ago
6 0
Where is the phone ? Can’t answer without a picture of the problem
elixir [45]3 years ago
4 0

The results illustrated here were hidden by means of Obscurational Cloaking technology.

This technology is used to hide, conceal, obscure, and withhold results.

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What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally h
zysi [14]

Complete question:

What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)

Answer:

The peak emf generated by the coil is 15.721 kV

Explanation:

Given;

Radius of coil, r = 0.250 m

Number of turns, N = 500-turn

time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s

magnetic field strength, B = 0.425 T

Induced peak emf = NABω

where;

A is the area of the coil

A = πr²

ω is angular velocity

ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s =  60 rev/s

Induced peak emf = NABω

                               = 500 x (π x 0.25²) x 0.425 x 376.738

                               = 15721.16 V

                               = 15.721 kV

Therefore, the peak emf generated by the coil is 15.721 kV

5 0
3 years ago
A circular plate of 500-mm diameter is maintained at T1 = 600 K and is positioned coaxial to a conical shape. The back side of t
drek231 [11]
For part a)
Since the conical surface is not exposed to the radiation coming from the walls only from the circular plate and assuming steady state, the temperature of the conical surface is also equal to the temperature of the circular plate. T2 = 600 K

For part b)
To maintain the temperature of the circular plate, the power required would be calculated using:
Q = Aσ(T₁⁴ - Tw⁴)
Q = π(500x10^-3)²/4 (5.67x10^-8)(600⁴ - 300⁴)
Q = 5410.65 W
5 0
3 years ago
Can someone please help me ASAP please!
natulia [17]
It would be D (hoping the picture you posted is what you needed help with
8 0
3 years ago
___ charges attract each other
klio [65]

Answer:

Opposite charges attract each other.

6 0
3 years ago
Read 2 more answers
How does the de broglie wavelength of an electron change if its momentum increases?
Virty [35]

Answer:

The De Broglie wavelength decreases

Explanation:

The relationship between the De Broglie wavelength of a particle and its momentum is given by

\lambda=\frac{h}{p}

where

\lambda is the De Broglie wavelength of the particle

h is the Planck constant

p is the momentum of the particle

As we see from the formula, there is an inverse relationship between the De Broglie's wavelength and the momentum. Therefore, we can conclude that:

- if the momentum of the electron increases,

- its De Broglie wavelength will decrease

and vice-versa.

3 0
3 years ago
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