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NNADVOKAT [17]
3 years ago
10

PLEASE HELP!!

Physics
1 answer:
adelina 88 [10]3 years ago
6 0

Answer:

D

Explanation:

Michael Faraday is probably best known for his discovery of electromagnetic induction, his contributions to electrical engineering and electrochemistry or due to the fact that he was responsible for introducing the concept of field in physics to describe electromagnetic interaction.

Electromagnetic or magnetic induction is the production of an electromotive force across an electrical conductor in a changing magnetic field.

Electrical engineering is an engineering discipline concerned with the study, design and application of equipment, devices and systems which use electricity, electronics, and electromagnetism.

Electrochemistry is the branch of physical chemistry that studies the relationship between electricity, as a measurable and quantitative phenomenon, and identifiable chemical change, with either electricity considered an outcome of a particular chemical change or vice versa.

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A bag of cement weighing 325 N hangs in equilibrium from three wires. Two of the wires make angles of theta1=60.0 degrees and th
Murljashka [212]
The sketch of the system is: two strings, 1 and 2, are attached to the ceiling and to a third string, 3.The third string holds the bag of cement. 

The free body diagram of the weight with the string 3, drives to the tension T3 = weihgt => T3 = 325 N

The other free body diagram is around the joint of the three strings.

In this case, you can do the horizontal forces equilibrium equation as:

T1* cos(60) - T2*cos(40) = 0

And the vertical forces equilibrium equation:

Ti sin(60) + T2 sin(40) = T3 = 325 N

Then you have two equations with two unknown variables, T1 and T2

0.5 T1 - 0.766 T2 = 0

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When you solve it you get, T1 = 252.8 N and T2 = 165 N

Answer: T1 = 252.8 N, T2 = 165N, and T3 = 325N



 
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3 years ago
What is the voltage across a semiconductor bar if the current through it is 0.17 A? The electron concentration in the bar is 2.7
Anastaziya [24]

Answer:

The voltage across a semiconductor bar is 0.068 V.

Explanation:

Given that,

Current = 0.17 A

Electron concentration n= 2.7\times10^{18}\ cm^{-3}

Electron mobility \mu=1000 cm^2/Vs

Length = 0.1 mm

Area = 500 μm²

We need to calculate the resistivity

Using formula of resistivity

\sigma=n\times q\times \mu

\rho=\dfrac{1}{\sigma}

Put the value into the formula

\rho=\dfrac{1}{2.7\times10^{18}\times10^{6}\times1.6\times10^{-19}\times1000\times10^{-4}}

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Using formula of resistance

R=\dfrac{\rho l}{A}

R=\dfrac{2\times10^{-6}\times0.1\times10^{-3}}{500\times(10^{-6})^2}

R=0.4\ \Omega

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Using formula of voltage

V= IR

Put the value into the formula

V=0.17\times0.4

V=0.068\ V

Hence, The voltage across a semiconductor bar is 0.068 V.

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B. is the answer.

C is not correct because the light is actually reflected off of an opaque object.
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