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2 years ago

Light of frequency 2.5 x 1015 Hz illuminates a

Physics

1 answer:

Nikitich [7]2 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

PLEASE HELP ME WITH THIS PROBLEM

valentinak56 [21]

1) The mass of the continent is $2.13\cdot 10^{21} kg$

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3) The speed of the jogger must be 2.76 m/s

Explanation:

The continent is a slab of side 5900 km (so the surface is 5900 x 5900, assuming it is a square) and depth 26 km, therefore its volume is:

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The mass of the continent is given by

$m=\rho V$

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$\rho = 2790 kg/m^3$ is its density

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$m=(2790)(7.62\cdot 10^{17})=2.13\cdot 10^{21} kg$

To find the kinetic energy, we need to convert the speed of the continent into m/s first.

The speed is

v = 1.6 cm/year

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1.6 cm = 0.016 m

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So, the speed is

$v=\frac{0.016 m}{3.15 \cdot 10^7 s}=5.08\cdot 10^{-10}m/s$

Now we can find the kinetic energy of the continent, which is given by

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

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$m=2.13\cdot 10^{21} kg$ is the mass

$v=5.08\cdot 10^{-10}m/s$ is the speed

Substituting,

$K=\frac{1}{2}(2.13\cdot 10^{21})(5.08\cdot 10^{-10})^2=274.8 J$

The jogger in this part has the same kinetic energy of the continent, so

K = 274.8 J

And its mass is

m = 72 kg

We can write his kinetic energy as

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

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And solving the equation for v, we find his speed:

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