7: A 2 mA current passes through a 1.4 cm long solenoid producing a magnetic field of .162 G. How many turns are in the solenoid

Answers

Answer 1

Answer:

Answer:

The number of turns is 64449395

Explanation:

The expression for the solenoid formula is stated below, and it is what we are going to use to solve for the number of turns

B= μ₀nl

where B= magnetic field

μ₀= permeability , 4π × 10⁻⁷ Henry

n= number of turns

l= length of coil

Given data

current I= 2 mA

length L= 1.4 cm to meter we have 0.014

magnetic field B= 0.162 T

From the expression we can make n subject of formula we have

n=B/μ₀l

Substituting we have

n= 0.162/4π × 10⁻⁷*2*10^-3

n= 0.162/8π × 10^⁻10

n= 0.162/2.5136*10^-9

n= 64449395

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2. Annealed low-carbon steel has a flow curve with strength coefficient of 75000 psi and strain-hardening exponent of 0.25. A tensile test specimen with a gauge length of 2 in. is stretched to a length of 3.3 in. Determine the flow stress and the average flow stress that the metal experienced during this deformation.

Answers

Answer:

Flow stress= 9390Psi

Average flow stress= 4173.33Psi

Explanation:

Given:

Strength Coefficient = 75000psi

Strain hardening Exponent = 0.25

Gauge length = 2inches

Stretch length = 3.3 inches

Flow stress,Yf = 75000 × ln(3.3/2) × 0.25

Yf = 75000× ln(1.65) × 0.25

Yf = 75000× 0.5008 × 0.25

Flow stress = 9390Psi

Average flow stress = 75000× 0.5008 × (0.25/2.25)

Average flow stress= 4173.33psi

A golf ball is dropped from rest from a height of 9.50 m. It hits the pavement, then bounces back up, rising just 9.70 m before falling back down again. A boy catches the ball on the way down when it is 1.20 m above the pavement. Ignoring air resistance calculate the total amount of time the ball is in the air, from drop to catch?

Answers

The ball is in the air for 27.70m because
9.5+9.7=19.2...
9.7-1.2=8.5...
19.2+8.5=27.70...

Why are chemical processes unable to produce the same amount of energy flowing out of the sun as nuclear fusion?

Answers

Answer:

Explanation:

One of the major differences between nuclear reactions and chemical reactions is that nuclear reactions involve larger amount of energy than chemical energy. This is because the force between the protons and neutrons in the nucleus of an atom is much higher than the force of attraction between electrons and the positively charged nucleus, hence nuclear reactions involves/requires a larger amount of energy (because it's reactions involve the nucleus) than chemical reactions (because it's reactions involve the electrons).

Thus, during nuclear fusion, two light nuclei are bombarded against one another to produce a larger/heavier nuclei with the release of large amount of energy (because the forces between the protons and neutrons are much higher) unlike when two atoms/molecules are chemically combined together to form a new molecule with the rearrangement of electrons in the valence shells of the participating molecules.

The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 1025 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 55 volts?

Answers

Explanation:

Given data:

Area A = 10 cm×2 cm = 20×10⁻⁴ m²

Distance d between the plates = 1 mm = 1×10⁻³m

Voltage of the battery is emf = 100 V

Resistance = 1025 ohm

Solution:

In RC circuit, the voltage between the plates is related to time t. Initially the voltage is equal to that of battery V₀ = emf = 100V. But After time t the resistance and capacitor changes it and the final voltage is V that is given by

$V = V_(0)(1-e^{(-t)/(RC) } )\n(V)/(V_(0) ) = 1-e(^{(-t)/(RC) }) \ne^{(-t)/(RC) } = 1- (V)/(V_(0) )$

Taking natural log on both sides,

$e^{(-t)/(RC) } = 1- (V)/(V_(0) ) \n(-t)/(RC) = ln(1-(V)/(V_(0) ) )\nt = -RCln(1 - (V)/(V_(0) ))$

$t = -RC ln (1-(V)/(V_(0) ))$ (1)

Now we can calculate the capacitance by using the area of the plates.

C = ε₀A/d

= $((8.85*10^(-12))) (20*10^(-4)) )/(1*10^(-3) )$

= 18×10⁻¹²F

Now we can get the time when the voltage drop from 100 to 55 V by putting the values of C, V₀, V and R in the equation (1)

$t = -RC ln (1-(V)/(V_(0) ))$

= -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)

= 15×10⁻⁹s

= 15 ns

"For the lowest harmonic of pipe with two open ends, how much of a wavelength fits into the pipe’s length?"

Answers

Answer:

0.5 lambda(wavelength)

Explanation:

We know that

The first harmonic for both side open ended pipe is

L= 1/2lambda

So L = 0.5*wavelength

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