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3 years ago

What is the acceleration of a runner who increases her velocity from 2m/s to 8m/s in 3s?

Physics

2 answers:

Tomtit [17]3 years ago

8 0

Answer: 2 m/s^2

Explanation:

V1-Vf/ T

2m/s - 8m/s / 3s

6m/s / 3s =

2 m/s^2

borishaifa [10]3 years ago

3 0

initial velocity=u=2m/s
Final velocity=v=8m/s
Time=t=3s

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{8-2}{3}$

$\\ \sf\longmapsto Acceleration=\dfrac{6}{3}$

$\\ \sf\longmapsto Acceleration=2m/s^2$

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3 years ago

Calculate the intensity of thermal radiation from a blackbody by using Raleigh-Jeans Law at a temperature of 1000 K and with a w

ch4aika [34]

Explanation:

Given that,

Temperature = 1000 K

We need to calculate the intensity of thermal radiation from a black body

For wavelength 1000 nm

Using Raleigh-Jeans Law

$B_{k}=\dfrac{8\pi kT}{\lambda^4}$

Put the value into the formula

$B_{k}=\dfrac{8\times\pi\times1.38\times10^{-23}\times1000}{(1000\times10^{-9})^{4}}$

$B_{k}=346831.828956$

$B_{k}=34.68\times10^{4}\ W$

For wavelength 500 nm

$B_{k}=\dfrac{8\times\pi\times1.38\times10^{-23}\times1000}{(500\times10^{-9})^{4}}$

$B_{k}=5549309$

$B_{k}=55.49\times10^{5}\ W$

For wavelength 100 nm

$B_{k}=\dfrac{8\times\pi\times1.38\times10^{-23}\times1000}{(100\times10^{-9})^{4}}$

$B_{k}=3.47\times10^{9}\ W$

We need to calculate the intensity

Using Plank's Law

$E=\dfrac{2hc^2}{\lambda^5}\times\dfrac{1}{e^{\dfrac{hc}{\lambda k T}}-1}$

For wavelength 1000 nm

Put the value into the formula

$E=\dfrac{2\times6.63\times10^{-34}\times(3\times10^{8})^2}{(1000\times10^{-9})^5}\times\dfrac{1}{e^{\dfrac{6.63\times10^{-34}\times3\times10^{8}}{1000\times10^{-9}\times 1.38\times10^{-23}\times1000}}-1}$

$E=65.65\times10^{6}\ W/m^2.K^4$

For wavelength 500 nm

$E=\dfrac{2\times6.63\times10^{-34}\times(3\times10^{8})^2}{(500\times10^{-9})^5}\times\dfrac{1}{e^{\dfrac{6.63\times10^{-34}\times3\times10^{8}}{500\times10^{-9}\times 1.38\times10^{-23}\times1000}}-1}$

$E=11.56\times10^{2}\ W/m^2.K^4$

For wavelength 100 nm

$E=\dfrac{2\times6.63\times10^{-34}\times(3\times10^{8})^2}{(100\times10^{-9})^5}\times\dfrac{1}{e^{\dfrac{6.63\times10^{-34}\times3\times10^{8}}{100\times10^{-9}\times 1.38\times10^{-23}\times1000}}-1}$

$E=3.032\times10^{-44}\ W/m^2.K^4$

Hence, This is the required solution.

8 0

3 years ago