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Paladinen [302]

3 years ago

12

The universe is filled with photons left over from the Big Bang that today have an average energy of about 4.9 ✕10-4 (correspond

ing to a temperature of 2.7 K).
What is the number of available energy states per unit volume for these photons in an interval of 4 ✕10-5eV?

1 answer:

motikmotik3 years ago

8 0

Answer:

The number of available energy is $4.820\times10^{45}$

Explanation:

Given that,

Energy $E=4.9\times10^{-4}\ J$

Temperature = 2.7 K

Energy states per unit volume $dE=4\times10^{-5}\ eV$

We need to calculate the number of available energy

Using formula of energy

$N=g(E)dE$

$N=\dfrac{8\pi\times E^2 dE}{(hc)^3}$

Where, h = Planck constant

c = speed of light

E = energy

Put the value into the formula

$N=\dfrac{8\pi\times(4.9\times10^{-4})^2\times4\times10^{-5}\times1.6\times10^{-19}}{(6.67\times10^{-34}\times3\times10^{8})^3}$

$N=4.820\times10^{45}$

Hence, The number of available energy is $4.820\times10^{45}$

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At one point in a pipeline, the water's speed is 3.57 m/s and the gauge pressure is 68.7 kPa. Find the gauge pressure at a secon

ArbitrLikvidat [17]

Answer:

The pressure at point 2 is $P_2 = 254.01 kPa$

Explanation:

From the question we are told that

The speed at point 1 is $v_1 = 3.57 \ m/s$

The gauge pressure at point 1 is $P_1 = 68.7kPa = 68.7*10^{3}\ Pa$

The density of water is $\rho = 1000 \ kg/m^3$

Let the height at point 1 be $h_1$ then the height at point two will be

$h_2 = h_1 - 18.5$

Let the diameter at point 1 be $d_1$ then the diameter at point two will be

$d_2 = 2 * d_1$

Now the continuity equation is mathematically represented as

$A_1 v_1 = A_2 v_2$

Here $A_1 , A_2$ are the area at point 1 and 2

Now given that the are is directly proportional to the square of the diameter [i.e $A= \frac{\pi d^2}{4}$ ]

which can represent as

$A \ \ \alpha \ \ d^2$

=> $A = c d^2$

where c is a constant

so $\frac{A_1}{d_1^2} = \frac{A_2}{d_2^2}$

=> $\frac{A_1}{d_1^2} = \frac{A_2}{4d_1^2}$

=> $A_2 = 4 A_1$

Now from the continuity equation

$A_1 v_1 = 4 A_1 v_2$

=> $v_2 = \frac{v_1}{4}$

=> $v_2 = \frac{3.57}{4}$

$v_2 = 0.893 \ m/s$

Generally the Bernoulli equation is mathematically represented as

$P_1 + \frac{1}{2} \rho v_1^2 + \rho * g * h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho * g * h_2$

So

$P_2 = \rho * g (h_1 -h_2 )+P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )$

=> $P_2 = \rho * g (h_1 -(h_1 -18.3) + P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )$

substituting values

$P_2 = 1000 * 9.8 (18.3) )+ 68.7*10^{3} + \frac{1}{2} * 1000 ((3.57)^2 -0.893 ^2 )$

$P_2 = 254.01 kPa$

8 0

3 years ago

A uniform, spherical, 1900.0 kg shell has a radius of 5.00 m. Find the gravitational force this shell exerts on a 1.80 kg point

Mandarinka [93]

Answer:

$F=9.09\times 10^{-9} N$

Explanation:

We are given that

Mass of spherical shell, $m_1$ =1900 kg

Mass= $m_2=1.80 kg$

Radius of shell=r=5 m

Distance between two masses=r=5.01 m

Because distance measure from center .

Gravitational force

$F=G\frac{m_1m_2}{r^2}$

$G=6.67\times 10^{-11} Nm^2/kg^2$

Using the formula

$F=6.67\times 10^{-11}\times \frac{1900\times 1.80}{(5.01)^2}$

$F=9.09\times 10^{-9} N$

Hence,the gravitational force = $F=9.09\times 10^{-9} N$

6 0

3 years ago

Suppose that the height of the incline is h = 14.7 m. Find the speed at the bottom for each of the following objects. 1.solid sp

tensa zangetsu [6.8K]

Answer:

1. 14.4 m/s 2. 13.2 m/s 3. 12.0 m/s 4. 13.9 m/s

Explanation:

Assuming no friction present, the different objects roll without slipping, so there is a constant relationship between linear and angular velocity, as follows:

ω= v/r

If no friction exists, the change in total kinetic energy must be equal in magnitude to the change in the gravitational potential energy:

∆K = -∆U

½ *m*v² + ½* I* ω² = m*g*h

Simplifying and replacing the value of the angular velocity:

½ * v² + ½ I *(v/r)² = g*h (1)

In order to answer the question, we just need to replace h by the value given, and I (moment of inertia) for the value for each different object, as follows:

Solid Sphere I = 2/5* m *r²

Replacing in (1):

½ * v² + ½ (2/5 *m*r²) *(v/r)² = g*h

Replacing by the value given for h, and solving for v:

v = √(10/7*9.8 m/s2*14.7 m) = 14. 4 m/s

Spherical shell I=2/3*m*r²

Replacing in (1):

½ * v² + ½ (2/3 *m*r²) *(v/r)² = g*h

Replacing by the value given for h, and solving for v:

v = √(6/5*9.8 m/s2*14.7 m) = 13.2 m/s

Hoop I= m*r²

Replacing in (1):

½ * v² + ½ (m*r²) *(v/r)² = g*h

Replacing by the value given for h, and solving for v:

v = √(9.8 m/s2*14.7 m) = 12.0 m/s

Cylinder I = 1/2 * m* r²

Replacing in (1):

½ * v² + ½ (1/2 *m*r²) *(v/r)²= g*h

Replacing by the value given for h, and solving for v:

v = 2*√(1/3*9.8 m/s2*14.7 m) = 13.9 m/s

5 0

3 years ago

Which type of circuit is becoming more common in residential electrical design and construction?

jekas [21]

Communication circuit <em>(D)</em> is becoming more common in residential electrical design and construction.

LAN Ethernet cables, outlets, and even hubs and bridges, are being built into the walls of new homes, along with the usual electrical outlet wiring, to give the owner the networking infrastructure and internet access that everybody needs now ... without stringing a mess of cables on the floor and through doors all over the house.

3 0

3 years ago

What is the smallest lithostratigraphic rock unit?

goldenfox [79]

Answer:

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Explanation:

6 0

3 years ago

Read 2 more answers