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Bond [772]
3 years ago
12

In tin at room temperature, the mobility of mobile electrons is about 1.5 ✕ 10⁻³ (m/s)/(V/m), and there are about 3.7 ✕ 10²⁸ mob

ile electrons per m³. Calculate the conductivity σ. In actual practice, it is usually easier to measure the conductivity σ and deduce the mobility u from this measurement.
Physics
1 answer:
kifflom [539]3 years ago
6 0

Answer:

\sigma = 8.88 \times 10^6 \frac{1}{ohm-m}

Explanation:

As we know that mobility of electrons is given as

\mu = \frac{v_d}{E}

now we also know that

j = \sigma E

here we know

j = ne v_d

ne v_d = \sigma E

so we have

\sigma = ne \frac{v_d}{E}

\sigma = (3.7 \times 10^{28})(1.6 \times 10^{-19})(1.5 \times 10^{-3})

\sigma = 8.88 \times 10^6 \frac{1}{ohm-m}

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How many times did john glenn orbit the earth
galben [10]

Answer:

2 times

Explanation:

1st time on Feb. 20, 1962 at age 41

2nd time on Oct. 29, 1998 at age 77

8 0
3 years ago
A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A cons
Virty [35]

Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

H_m=1.65m

H_E=1.16307m

Explanation:

From the question we are told that

Mass of ball M=2kg

Length of string L= 2m

Wind force F=13.2N

Generally the equation for \angle \theta is mathematically given as

tan\theta=\frac{F}{mg}

\theta=tan^-^1\frac{F}{mg}

\theta=tan^-^1\frac{13.2}{2*2}

\theta=73.14\textdegree

Max angle =2*\theta= 2*73.14=>146.28\textdegree

Generally the equation for max Height H_m is mathematically given as

H_m=L(1-cos146.28)

H_m=0.9(1+0.8318)

H_m=1.65m

Generally the equation for Equilibrium Height H_E is mathematically given as

H_E=L(1-cos73.14)

H_E=0.9(1+0.2923)

H_E=1.16307m

8 0
2 years ago
The time for a sound wave to travel between two people is 0.80 s,
sasho [114]

Answer:

Explanation:

Using the below formula

Speed of sound = ( distance between observers) *2/(total time taken)

Now putt the given values ,

time taken = 0.80 sec

distance = 256 m

hence

V of sound= 256*2/0.80

V of sound = 640 m/sec

4 0
3 years ago
To what potential should you charge a 3.0 μf capacitor to store 1.0 j of energy?
Nimfa-mama [501]
The energy stored in a capacitor is given by:
U= \frac{1}{2}CV^2
where
U is the energy
C is the capacitance
V is the potential difference

The capacitor in this problem has capacitance
C=3.0 \mu F = 3.0 \cdot 10^{-6} F
So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
V= \sqrt{ \frac{2U}{C} }= \sqrt{ \frac{2 \cdot 1.0 J}{3.0 \cdot 10^{-6}F} }=816 V
8 0
3 years ago
Two equally charged, 1.00 g spheres are placed with 2.00 cm between their centers. when released, each begins to accelerate at 2
Leya [2.2K]
1) Force = m*a = 1.00 g * (1kg / 1000 g) * 225 m/s^2 = 0.225 N

2) Charge

Force = K (charge)^2 /(distance)^2 => charge = √ [Force * distance^2 / k]

k = 9.00 * 10^9 N*m^2 / C^2

charge = √ [0.225 N * (0.02 m)^2 / 9.00* 10^9 N*m^2 / C^2 ]

charge = 0.0000001 C = 0.0001 mili C
3 0
3 years ago
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