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Sergeeva-Olga [200]
2 years ago
13

Respostas 100 J 200 J 300 J 400 J 500 J

Physics
2 answers:
Aneli [31]2 years ago
7 0

Answer:

Explanation:

400 J

:> :)

Fiesta28 [93]2 years ago
5 0

400J

hope it's help

now is my exam and your

pls welcome to me

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A wagon wheel consists of 8 spokes of uniform diamter, each of mass m, and length L. The outer ring has a mass m rin. What is th
Genrish500 [490]

Answer:

L^2(\dfrac{8m}{3}+m_r)

Explanation:

m = Mass of each rod

L = Length of rod = Radius of ring

m_r = Mass of ring

Moment of inertia of a spoke

\dfrac{mL^2}{3}

For 8 spokes

8\dfrac{mL^2}{3}

Moment of inertia of ring

m_rL^2

Total moment of inertia

8\dfrac{mL^2}{3}+m_rL^2\\\Rightarrow L^2(\dfrac{8m}{3}+m_r)

The moment of inertia of the wheel through an axis through the center and perpendicular to the plane of the ring is L^2(\dfrac{8m}{3}+m_r).

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3 years ago
True or False A goal of humanist psychologists today is to help people reach their full potential.
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True, because people should reach their full potential within themselves and human psychologist should help those that need help to reach it.
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Nuclear energy is over ___ times stronger than the chemical bonds between the atoms
Kamila [148]

Answer:

1 millions times stronger

8 0
2 years ago
A ball is thrown straight up. At the top of its path its acceleration is
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C. Acceleration is the rate of change of velocity. So at the top of the path, while the velocity is zero, the CONSTANT GRAVITATIONAL ACCELERATION is about 10 m/s^2 (9.8)
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3 years ago
Ultraviolet light of wavelength 270 nm strikes a metal whose work function is 2.3 eV.What is the shortest de Broglie wavelength
soldier1979 [14.2K]

Answer:

The shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

Explanation:

Given;

wavelength of ultraviolet light, λ = 270 nm

work function of the metal, φ = 2.3 eV = 2.3 x 1.602 x 10⁻¹⁹ J = 3.685 x 10⁻¹⁹ J

The energy of the ultraviolet light is given by;

E = \frac{hc}{\lambda}\\\\E = \frac{(6.626*10^{-34} )(3*10^{8}) }{270*10^{-9} }\\\\E = 7.362 * 10^{-19} \ J

The energy of the incident light is related to kinetic energy of the electron and work function of the metal by the following equation;

E = φ  + K.E

K.E = E - φ

K.E = (7.362 x 10⁻¹⁹ J) - (3.685 x 10⁻¹⁹ J )

K.E = 3.677 x 10⁻¹⁹ J

K.E = ¹/₂mv²

mv² = 2K.E

velocity of the electron is given by;

V = \sqrt{\frac{2K.E}{m} }\\\\V =  \sqrt{\frac{2(3.677*10^{-19}) }{9.1*10^{-31} } }\\\\V = 8.99*10^{5}  \ m/s

the shortest de Broglie wavelength for the electrons is given by;

\lambda = \frac{h}{mv}\\ \\\lambda = \frac{6.626*10^{-34} }{(9.1*10^{-31})( 8.99*10^{5} )}\\\\\lambda = 8.10*10^{-10} \ m\\\\\lambda = 0.81 \ nm

Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

7 0
3 years ago
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