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3 years ago

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The temperature of the cosmic background radiation is measured to be 2.7 k. What is the wavelength of the peak in the spectral d

istribution at this temperature?

1 answer:

KATRIN_1 [288]3 years ago

7 0

Answer:

$1.07\cdot 10^{-3} m$

Explanation:

The peak wavelength of the spectral distribution can be found by using Wien's displacement law:

$\lambda=\frac{b}{T}$

where

$b=2.898\cdot 10^{-3} m\cdot K$ is Wien's displacement constant

T is the absolute temperature

For the cosmic background radiation, the temperature is

T = 2.7 K

So, the corresponding peak wavelength is

$\lambda=\frac{2.898\cdot 10^{-3} m\cdot K}{2.7 K}=1.07\cdot 10^{-3} m$

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Convert the speed of light 3.0x10^8 m/s to km/day

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Answer: $2.592 \times 10^{8}km/day$

$1 m = 0.001 km\\ 1 s= 1.157\times10^{-5} days\\ 1 m/s = \frac {0.001}{1.157\times10^{-5}} km/day = 86.4 km/day \\ 3.0\times 10^{8} m/s = 3.0\times 10^{8} m/s \times \frac {86.4 km/day}{1m/s} =2.592 \times 10^{8}km/day$

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Answer:

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Explanation:

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3 years ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

so

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

<u>Conservation of kinetic energy:</u>

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

so

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

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3 years ago

A transformer has 1000 turns in the primary coil and 100 turns in the secondary coil. If the primary coil is connected to a 120

alexdok [17]

Answer:

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Explanation:

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Using,

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Where Vs = Voltage in the secondary coil, Vp = Voltage in the primary coil, Ns = number of turn in the secondary coil, Np = number of turns in the primary coil.

Make Vs the subject of the equation

Vs = Vp(Ns/Np)........................ Equation 2

Given: Vp = 120 v, Np = 1000 turns, Ns = 100 turns

Substitute into equation 2

Vs = 120(100/1000)

Vs = 120×0.1

Vs = 12 v

For the current,

Using

Ns/Np = Ip/Is....................... Equation 3

Where Ip = current in the primary coil, Is = current in the secondary coil

make Is the subject of the equation

Is = Ip(Np/Ns).................. Equation 4

Given: Np = 1000 turns, Ns = 100 turns, Ip = 0.05 A

Substitute into equation 4

Is = 0.05(1000/100)

Is = 0.05×10

Is = 0.5 A.

Hence the voltage and the current in the secondary coil is 12 V, 0.5 A

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