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V125BC [204]
3 years ago
14

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?
Physics
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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3 years ago
Tyson Gay's best time to run 100.0 meters was 9.69 seconds. What was his average speed during this run, in miles per hour? (3.28
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Answer:

23.086 mile/h

Explanation:

Given,

Distance Tyson Gay run = 100 m

time of run, t = 9.69 s

average speed of the in mph = ?

Speed of the Gay = \dfrac{distance}{time}

v = \dfrac{100}{9.69}

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4 0
2 years ago
Read 2 more answers
A wire 3.22 m long and 7.32 mm in diameter has a resistance of 11.9 mΩ. A potential difference of 33.7 V is applied between the
Scorpion4ik [409]

Answer:

(a) Current is 2831.93 A

(b) 8.40A/m^2

(c) \rho =15.52\times 10^{-9}ohm-m

Explanation:

Length of wire l = 3.22 m

Diameter of wire d = 7.32 mm = 0.00732 m

Cross sectional area of wire

A=\pi r^2=3.14\times 0.00366^2=4.20\times 10^{-5}m^2

Resistance R=11.9mohm=11.9\times 10^{-3}ohm

Potential difference V = 33.7 volt

(A) current is equal to

i=\frac{V}{R}=\frac{33.7}{11.9\times 10^{-3}}=2831.93A

(B) Current density is equal to

J=\frac{i}{A}

J=\frac{2831.93}{4.20\times 10^{-5}}=8.40A/m^2

(c) Resistance is equal to

R=\frac{\rho l}{A}

11.9\times 10^{-3}=\frac{\rho \times 3.22}{4.20\times 10^{-5}}

\rho =15.52\times 10^{-9}ohm-m

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