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Y_Kistochka [10]

1 year ago

6

A 64kg wheel with radius 2.4m is rotating at 280rev/min. if you want to stop it in 30sec. (use point mass of moment of inertia)

A. how much work must be done to stop it? B. what is the required power?

1 answer:

AfilCa [17]1 year ago

4 0

Answer:

Alot of work must be done

Im not sure what the required power will be

Explanation:

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A car is initially traveling at 12 m/s when the driver sees a yellow light ahead. He accelerates at a constant 7 m/s^2 for 6 s i

Vitek1552 [10]

Answer:

v = 54m/s

Explanation:

$a = \frac{v - u}{t}$

a = 7m/s²

u = 12m/s

t = 6s

7 = (v-12)/6

v - 12 = 42

v = 54m/s

(Correct me if i am wrong)

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2 years ago

Which law states that absolute zero cannot be reached?

Lera25 [3.4K]

Assuming you mean temperature

Answer: The third law of thermodynamics

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2 years ago

A person drives for 1.00 h at 10.0 km/h and 2.00 h at 20.0 km/h. What is their average speed, in SI units?

Veronika [31]

I believe it is b if i’m correct

6 0

2 years ago

X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

$\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)$

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

$\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m$

b) The change in photon energy is given by:

$\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm$

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

$P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\$

$E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV$

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