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Alika [10]
1 year ago
9

All of the following should be considered when designing a study except __________.

Physics
1 answer:
kicyunya [14]1 year ago
8 0

Answer:

A the type of research method to use

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A 3" diameter germanium wafer that is 0.020" thick at 300K has 1.015 x 10^17 As atoms added to it. What is the resistivity of th
Ber [7]

Answer:

0.546 ohm / μm

Explanation:

Given that :

N = 1.015 * 10^17

Electron mobility, u = 3900

Hole mobility, h = 1900

Ng = 4.42 x10^22

q = 1.6*10^-19

Resistivity = 1/qNu

Resistivsity (R) = 1/(1.6*10^-19 * 1.015 * 10^17 * 3900)

= 0.01578880889 ohm /cm

Resistivity of germanium :

R = 1 / 2q * sqrt(Ng) * sqrt(u*h)

R = 1 / 2 * 1.6*10^-19 * sqrt(4.42 x10^22) * sqrt(3900*1900)

R = 1 /0.0001831

R = 5461.4964 ohm /cm

5461.4964 / 10000

0.546 ohm / μm

7 0
2 years ago
Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's m
Reil [10]

Answer:

\Delta l=0.015m

Explanation:

We have given initial length of the steel guitar l = 1 m

Cross sectional area A=0.5mm^2=0.5\times 10^{-6}m^2

Young's modulus \gamma=2\times 10^{11}Pa

Force F = 1500 N

So stress =\frac{force}{area}=\frac{1500}{0.5\times 10^{-6}}=3000\times 10^{-6}=3\times 10^{9}Pa

We know that young's modulus =\frac{stress}{strain}

So 2\times 10^{11}=\frac{3\times 10^{9}}{strain}

strain=1.5\times 10^{-2}=0.015m

Now strain =\frac{\Delta l}{l}

0.015=\frac{\Delta l}{1}

\Delta l=0.015m

6 0
3 years ago
Read 2 more answers
A student hangs a wood block from a spring. The student pulls the block downwards so the spring is stretched and holds the block
Murrr4er [49]
Kinetic energy because of the wood plank is just still and not moving it’s potential but since it’s asking which one it doesn’t have it doesn’t have kinetic energy cause it’s not moving
6 0
3 years ago
In which of these examples is the greatest movement occurring?
vlada-n [284]
You need to provide a picture or tell us the examples... we can’t see what you see
7 0
3 years ago
Read 2 more answers
A conductor carrying a current I = 16.5 A is directed along the positive x axis and perpendicular to a uniform magnetic field. A
Jet001 [13]

To solve this problem we will apply the concepts related to the Magnetic Force, this is given by the product between the current, the body length, the magnetic field and the angle between the force and the magnetic field, mathematically that is,

F = ILBsin \theta

Here,

I = Current

L = Length

B = Magnetic Field

\theta = Angle between Force and Magnetic Field

But \theta = 90\°

F = ILB

Rearranging to find the Magnetic Field,

B = \frac{F}{IL}

Here the force per unit length,

B = \frac{1}{I}\frac{F}{L}

Replacing with our values,

B = \frac{0.130N/m}{16.5}

B = 0.0078T

Therefore the magnitude of the magnetic field in the region through which the current passes is 0.0078T

6 0
3 years ago
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