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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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Substance 1 because it became a liquid faster

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2 years ago
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100 meters toward the shore in 25 seconds
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Answer:

The speed of a turtle is 4m/s

Explanation:

speed = distance/time

distance = 100m

time = 25s

s = 100/25

= 4m/s

6 0
3 years ago
A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.480 m/s. The to
SpyIntel [72]
This is a problem of conservation of momentum

Momentum before throwing the rock: m*V = 96.0 kg * 0.480 m/s = 46.08 N*s

A) man throws the rock forward

=>

rock:
m1 = 0.310 kg
V1 = 14.5 m/s, in the same direction of the sled with the man

sled and man:
m2 = 96 kg - 0.310 kg = 95.69 kg
v2 = ?

Conservation of momentum:
momentum before throw = momentum after throw

46.08N*s = 0.310kg*14.5m/s + 95.69kg*v2

=> v2 = [46.08 N*s - 0.310*14.5N*s ] / 95.69 kg = 0.434 m/s

B) man throws the rock backward

this changes the sign of the velocity, v2 = -14.5 m/s

 46.08N*s = - 0.310kg*14.5m/s + 95.69kg*v2

v2 = [46.08 N*s + 0.310*14.5 N*s] / 95.69 k = 0.529 m/s


3 0
3 years ago
A thin walled spherical shell is rolling on a surface. What
ExtremeBDS [4]

Answer:

=\frac{1/3}{5/6} = 0.4

Explanation:

Moment of inertia of given shell= \frac{2}{3} MR^2

where

M represent sphere mass

R -sphere radius

we know linear speed is given as v = r\omega

translational K.E = \frac{1}{2} mv^2 = \frac{1}{2} m(r\omega)^2

rotational K.E = \frac{1}{2} I \omega^2 = \frac{1}{2} \frac{2}{3} MR^2 \omega^2

total kinetic energy will be

K.E = \frac{1}{2} m(r\omega)^2 + \frac{1}{2} \frac{2}{3} MR^2 \omega^2

K.E =\frac{5}{6} MR^2 \omega^2

fraction of rotaional to total K.E

=\frac{1/3}{5/6} = 0.4

8 0
3 years ago
Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400
hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

v^2=u^2+2as

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

v=u+at

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m

The beetle can jump to a height of <u>2.1 m</u>



7 0
3 years ago
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