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PolarNik [594]
3 years ago
10

If you drop a ball off a cliff, it starts out a 0 m/s. After 1 s, it will be traveling at about 10 m/s. If air resistance is rem

oved, what will happen in 2 s? And how do you know this?
A. The ball will still be moving at 10 m/s.
B. The ball will accelerate to 10 m/s.
C. The ball will accelerate to about 20 m/s.
Physics
2 answers:
Softa [21]3 years ago
8 0

Answer: (C) The ball will accelerate to about 20 m/s.

Explanation: The ball is in free fall and no other but gravitational force acts on it. This means the ball will be subject to the gravitational acceleration, which is about 9.8 m/s^2. So after one additional second the ball will increase its velocity to (10+9.8) m/s = 19.8 m/s or approx. 20 m/s.

ziro4ka [17]3 years ago
3 0
This anwser is C !!!!
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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

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Answer:

B- The amount of matter in a certain

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