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IRISSAK [1]
3 years ago
6

Modern physics theories indicate that:

Physics
2 answers:
Aleonysh [2.5K]3 years ago
7 0

Answer:

A) All particles exhibit wave behavior

Explanation:

According to Modern physics,  which may also be called Quantum Physics, all matter  exhibits both wave like and particle like character.

Light and other electromagnetic waves also exhibits both wave like and particle like behavior.This is called wave particle duality.

For gross objects wave length is negligible and so they cannot be identified  as wave. But every matter behaves as a wave as well as a particle.

Compton effect gives the best example of the wave particle duality.

vitfil [10]3 years ago
6 0

Answer:

B. Only moving particles exhibit wave behavior.

Explanation:

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Compare the magnitude of the magnetic field at the center of a circular current loop of radius 30 mm with the magnitude of the m
NISA [10]

Answer:

Ratio of magnetic field will be \frac{B}{B'}=0.0055

Explanation:

We have given radius of the loop r = 30 mm = 0.03 m

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We know that magnetic field due to solenoid is given by

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Now dividing eqn 1 by eqn 2

\frac{B}{B'}=\frac{\frac{\mu _0i}{2r}}{\mu _0ni}=\frac{1}{2nr}=\frac{1}{2\times 3000\times 0.03}=0.0055

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4 years ago
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and
nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}

k = 1.4

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Work done is given as;

W = \frac{1}{2} *m*(v_i^2 - v_e^2)

inlet velocity is negligible;

v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41  \ m/s

Therefore, the exit velocity is 629.41 m/s

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3 years ago
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GalinKa [24]

Explanation:

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