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Svetradugi [14.3K]
3 years ago
8

Lead atoms occupy a volume of 3 x 10-29 m3. Each atom contributes two free electrons. Calculate the Fermi velocity of lead.

Physics
1 answer:
Juliette [100K]3 years ago
4 0

<u>Answer:</u> The Fermi velocity of lead is 64.4 km/s.

<u>Explanation:</u>

To calculate the Fermi velocity, we use the equation:

V_f=\frac{h}{2\pi m_e}(\frac{3\pi^2N}{V})^{1/3}

where,

h = Planck's constant = 6.62\times 10^{-34}Js

m_e = mass of electron = 9.1\times 10^{-31}kg

N = Number of atoms present in per volume of atom multiplied by number of electrons present in given atom = \frac{2\times N_A\times V}{M}

N_A = Avogadro's number = 6.022\times 10^{26}mol^{-1}    (When the mass is in kilograms)

V = Volume = 3\times 10^{-29}m^3

M = molecular weight of lead = 207.2 g/mol

Putting values in above equation, we get:

V_f=\frac{6.62\times 10^{-34}}{2\times 3.14\times (9.1\times 10^{-31})}(\frac{3\times (3.14)^2\times (2\times 6.022\times 10^{23}\times 3\times 10^{-29})}{3\times 10^{-29}\times 207.2})^{1/3}

V_f=0.0644\times 10^6m/s=64.4km/s     (Conversion factor: 1 km = 1000 m)

Hence, the Fermi velocity of lead is 64.4 km/s

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Is 0.0 m/s squared increasing or decreasing
Agata [3.3K]
Decrease because its 0.0 m/s
4 0
3 years ago
A girl rolls a ball up an incline and allows it to re- turn to her. For the angle and ball involved, the acceleration of the bal
zalisa [80]

Answer:

3.28 m

3.28 s

Explanation:

We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.

Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

V0 = 4 m/s

a = -2.45 m/s^2 (because the acceleration is down slope)

Then:

X(t) = 4*t - 1.22*t^2

And the equation for speed is:

V(t) = V0 + a * t

V(t) = 4 - 2.45 * t

If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:

0 = 4 - 2.45 * t

4 = 2.45 * t

t = 1.63 s

Replacing that time on the position equation:

X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m

To find the time it will take to return we equate the position equation to zero:

0 = 4 * t - 1.22 * t^2

Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:

0 = t * (4 - 1.22*t)

t1 = 0

0 = 4 - 1.22*t2

1.22 * t2 = 4

t2 = 3.28 s

7 0
2 years ago
A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit
inn [45]

Answer:twice of initial value

Explanation:

Given

spring compresses x_1 distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1

When speed doubles

\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2

divide 1 and 2

\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}

x_2=2x_1

Therefore spring compresses twice the initial value

   

7 0
2 years ago
Hurry
marishachu [46]

6x8 = 48 feet

you can jump 48 feet on the moon

6 0
2 years ago
A disk-shaped grindstone of mass 3.0 kg and radius 8.0 cm is spinning at 600 rev/min. After the power is shut off, a man continu
kolbaska11 [484]

Answer:

τ=0.060 N.m

Explanation:

By kinematics:

\omega f = \omega o-\alpha*t

Solving for α:

\alpha=\frac{\omega o-\omega f}{t}

where ωo = 600*2*π/60;   ωf = 0;    t=10s

\alpha=6.283rad/s^2

The sum of torque is:

\tau=I*\alpha

\tau=M*R^2/2*\alpha

\tau=0.060 N.m

8 0
3 years ago
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