How far can you go until you reach space?

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

3 years ago

A 75 A resistor in a circuit has a current flowing through it of 2.0 A. What is

Reptile [31]

Answer:

The power dissipated by the resistor can be calculated as $P = I^{2} R$ .

Explanation:

Joules heating law states that any conductor passed with the electric current produces heating effect which is the electric power loss.

The given circuit has, $R = 75 ohm$ resistor (the unit is wrong i question because resistor's unit is ohm not Ampere).
Current flowing through the circuit is $I = 2.0 A$ .

By using joules heating theory, $P = I^{2} R$ ,

$or,$ $P = 2^{2} * 75$

$or,$ $P= 4*75$

$or$ , $P= 300 W$ .

So the power dissipated by the resistor in the circuit is 300 W.

7 0

3 years ago

A thin flake of mica ( n = 1.58 ) is used to cover one slit ofa double slit interference arrangement. The central point on thevi

ELEN [110]

Answer:

the thickness of the mica is 6.64μm

Explanation:

By definition we know that the phase between two light waves that are traveling on different materials (in this case also two) is given by the equation

$\Phi = 2\pi(\frac{L}{\lambda}(n_1-n_2))$

Where

L = Thickness

n = Index of refraction of each material

$\lambda = Wavelength$

Our values are given as

$\Phi = 7(2\pi)L=tn_1 = 1.58n_2 = 1\lambda = 550nm$

Replacing our values at the previous equation we have

$\Phi = 2\pi(\frac{L}{\lambda}(n_1-n_2))7(2\pi) = 2\pi(\frac{t}{\lambda}(1.58-1))$

$t = \frac{7*550}{1.58-1}\\t = 6637.931nm \approx 6.64\mu m$

the thickness of the mica is 6.64μm

5 0

3 years ago

A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine

Tju [1.3M]

Answer:

$_T}=24.57Nm$

ω = 0.0347 rad/s²

a ≅ 1.07 m/s²

Explanation:

Given that:

mass of the model airplane = 0.741 kg

radius of the wire = 30.9 m

Force = 0.795 N

The torque produced by the net thrust about the center of the circle can be calculated as:

$_T } = Fr$

where;

F represent the magnitude of the thrust

r represent the radius of the wire

Since we have our parameters in set, the next thing to do is to replace it into the above formula;

So;

$_T}=(0.795)*(30.9)$

$_T}=24.57Nm$

(b)

Find the angular acceleration of the airplane when it is in level flight rad/s²

$_T}=I \omega$

where;

I = moment of inertia

ω = angular acceleration

The moment of inertia (I) can also be illustrated as:

$I = mr^2$

I = ( 0.741) × (30.9)²

I = 0.741 × 954.81

I = 707.51 Kg.m²

$_T}=I \omega$

Making angular acceleration the subject of the formula; we have;

$\omega = \frac{_T}{I}$

ω = $\frac{24.57}{707.51}$

ω = 0.0347 rad/s²

(c)

Find the linear acceleration of the airplane tangent to its flight path.m/s²

the linear acceleration (a) can be given as:

a = ωr

a = 0.0347 × 30.9

a = 1.07223 m/s²

a ≅ 1.07 m/s²

5 0

3 years ago

the impact time for a particular collision is 3.0 × 10-3 seconds and the impulse in the collision is 0.30 newton seconds. what i

Zanzabum

There are several information's already given in the question. The answer can be easily deduced using those information's.

Time = 3.0 * 10-3 seconds
Impulse = 0.30 newton
Then
Force = Impulse/Time
= 0.30/3.0 * 10-3
= 1 * 10^3 newtons.
I hope the above procedure is clear for you to understand and it has actually come to your great help.

3 0

4 years ago