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3 years ago

Light of frequency 2.5 x 1015 Hz illuminates a

Physics

1 answer:

Nikitich [7]3 years ago

3 0

Answer:

h f = W + KE

Input energy equals work function plus KE of emitted electron

W = 6.63E-34 * 2.5E15 - 6.3 * 1.6E-19

W = 6.63 * 2.5 * 10^-19 - 10.1 * E-19 ev (1ev = 1.6E-19 J)

W = (16.6 - 10.1)E-19 = 6.5E-19 J

h f = 6.5E-19 J for electrons to be emitted with zero KE

f = 6.5E-19 / 6.63E-34 = .98E-15 / sec = 9.8E-14 / sec (threshold)

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3 years ago

__________ is the school of psychology that believes perception is more than the sum of its parts—it involves a whole pattern. A

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3 years ago

Read 2 more answers

A ball rolls down the hill which has a vertical height of 15 m. Ignoring friction what would be the gravitational potential ener

trasher [3.6K]

a) Potential energy: 147 m [J]

The gravitational potential energy of an object is given by

$U=mgh$

where

m is its mass

$g=9.8 m/s^2$ is the acceleration of gravity

h is the height of the object above the ground

In this problem,

h = 15 m

We call 'm' the mass of the ball, since we don't know it

So, the potential energy of the ball at the top of the hill is

$U=(m)(9.8)(15)=147 m$ (J)

b) Velocity of the ball at the bottom of the hill: 17.1 m/s

According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:

$U=K=\frac{1}{2}mv^2$

where

v is the final velocity of the ball

We know from part a) that

U = 147 m

Substituting into the equation above,

$147 m = \frac{1}{2}mv^2$

And re-arranging for v, we find the velocity:

$v=\sqrt{2\cdot 147}=17.1 m/s$

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3 years ago

Make a

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3 years ago