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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?
Physics
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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On a Saturday afternoon, you decide to pay a neighborhood kid to mow your lawn. The kid usesa manual push lawn mower with a mass
Mars2501 [29]

Answer with Explanation:

We are given that

A.Mass,m=12 kg

\theta=53^{\circ}

\mu_k=0.16

Speed,v=1.5m/s

Net force in x direction must be zero

F_{net}=0

Fsin\theta-f=0

Fsin\theta=f

Net force in y direction

N-mg-Fcos\theta=0

N=mg+Fcos\theta

f=\mu_kN=\mu_k(mg+Fcos\theta)

Fsin\theta=\mu_k(mg+Fcos\theta)

Fsin\theta=\mu_kmg+\mu_kFcos\theta

Fsin\theta-\mu_kFcos\theta=\mu_kmg

F(sin\theta-\mu_kcos\theta)=\mu_kmg

F=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}

Power,P=Fv

P=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}v

Where g=9.8m/s^2

B.Substitute the values

P=\frac{0.16\times 12\times 9.8}{sin53-0.16cos53}\times 1.5

P=40.17W

6 0
3 years ago
A rock is dropped off a cliff and falls the first half of the distance to the ground in 2.0 seconds. how long will it take to fa
son4ous [18]

Answer:

The answer is 0.83 seconds.

Explanation:

The formula of free fall is following:

h=1/2*g*t^2

Where g=9.8 m/s^2 and t=2 seconds, the rock takes:

h=1/2*9.8*2^2=19.6

19.6 meters. This is the half distance of the cliff. The whole distance is 39.2 meters. So it takes:

39.2=1/2*9.8*t^2\\t^2=8\\t=2.83

2.83 second to fall down completely. The rock takes the second half of the cliff in 0.83 seconds

8 0
3 years ago
Soil can best be described as the
Rom4ik [11]
<span>The correct option is D. Soil can best be described as the loose covering of weathered rocks and decaying organic matter. There are different types of soil, the type of soil formed depends majorly on the type of parent rock from which the soil is formed and the amount of organic matter present in the soil.</span>
5 0
3 years ago
Write an expression for a transverse harmonic wave that has a wavelength of 2.5 m and propagates to the right with a speed of 13
lyudmila [28]

Answer:

y = 0.14 Cos\left ( 2.512x-34.66t \right )

Explanation:

wavelength, λ = 2.5 m

speed, v = 13.8 m/s

Amplitude, A = 0.14 m

The general equation of the transverse harmonic wave which is travelling right is given by

y = A Sin\left ( \frac{2\pi }{\lambda } (x - vt)+\phi \right  )

where, Ф is phase

At t = 0, x = 0 , y = 0.14 m

0.14 = 0.14 Sin Ф

Ф = π/2

So, the equation is

y = 0.14Sin\left ( \frac{2\pi }{2.5 } (x - 13.8t)+\frac{\pi }{2} \right  )

y = 0.14 Cos\left ( 2.512x-34.66t \right )

3 0
3 years ago
Calculate the approximate volume of a uranium nucleus, 23592u. (you can ignore the mass defect in this calculation and simply ta
tester [92]

Radius of nuclei is given by formula

R = R_oA^{1/3}

now we can say volume of the nuclei is given as

V = \frac{4}{3}\pi R_o^3* A

now the density is given as

density = mass / volume

mass of nuclei = mass of neutron + mass of protons

m = z*m_p + (A- z)*m_n

m_p = m_n = 1.008u

m = A*1.008u

Now density is given as

\rho = \frac{A*1.008u}{\frac{4}{3}\pi R_0^3* A}

here we know that

R_0 = 1.2 fm

\rho = \frac{1.008u}{\frac{4}{3}\pi*(1.2*10^{-15})^3}

\rho = 2.31 * 10^{17} kg/m^3

So from above we can say that density of all nuclei is almost same.

5 0
3 years ago
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