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4 years ago

Which describes the relationship between photon energy and the color of light?

Physics

2 answers:

kolbaska11 [484]4 years ago

4 0

Answer:

Red photons have the least amount of energy.

Explanation:

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Andrew [12]4 years ago

3 0

Photon energy is directly proportional to the frequency of electromagnetic radiation.
(That would also mean that it's inversely proportional to the wavelength.)

So the photon energy increases as you scan the chart of visible colors
moving from the red end of the rainbow to the blue end.

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A student has to work the following problem: A block is being pulled along at constant speed on a horizontal surface a distance

brilliants [131]

Answer:

The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.

Explanation:

Since block moves with constant speed

So, frictional force

f = FCosq

Work done by friction

W = - fd

W = - fd Cos q

The answer may be greater or less than - fdSinq. It depends on the value of q which is less than, or equal to 45°.

6 0

3 years ago

The wall is 500 sq. feet. A gallon of paint covers 160 sq. feet. What is an appropriate conversion factor to help determine how

lana66690 [7]

Answer:

Explanation:

Given

Wall is $500 ft^2$ in area

$160\ ft^2$ requires 1 gallon of paint

so using unitary method

$1\ ft^2 requires \frac{1}{160}$ gallon of paint

$500 ft^2$ wall will require $=500\times \frac{1}{160}$ gallons of paint

$=3.0125\ gallons$

4 0

3 years ago

calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end

sdas [7]

Answer: $0.8\ mm$

Explanation:

Given

length of wire $l=75\ cm$

change in length $\Delta l=1.85\ mm$

mass of wire $m=10\ kg$

Young's modulus for silver $E=7.9\times 10^{10}\ N/m^2$

load on wire $F=mg$

$F=10\times 9.8=98\ kg$

change in length is given by

$\Delta l=\dfrac{Pl}{AE}$

Where A=area of cross-section

$A=\dfrac{Pl}{\Delta lE}$

$A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}$

$A=\dfrac{73.5}{14.615\times 10^{7}}$

$A=5.029\times 10^{-7}\ m^2$

also wire is the shape of cylinder so cross-section is given by

$A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2$

$\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }$

$\Rightarrow d^2=64.02\times 10^{-8}$

$\Rightarrow d=8\times 10^{-4}\ m$

$\Rightarrow d=0.8\ mm$

4 0

3 years ago

Which statement is true of a concave lens?

natka813 [3]

It produces only virtual images is the answer

5 0

3 years ago

Read 2 more answers

A 7.0 mm -diameter copper ball is charged to 40 nC. What fraction of its electrons have been removed? The density of copper is 8

mylen [45]

Answer:

$f = 2.6 \times 10^{-13}$

Explanation:

Let the mass of copper ball is "m" gram

now the total number of copper atom present in the ball is given as

$N = \frac{m}{29} \times 6.02 \times 10^{23}$

now the total number of electrons in one copper atom is 29

so total number of electrons in given sample of copper ball is

$N_e = m(6.02 \times 10^{29})$

now diameter of the ball is 7.0 mm

density of the ball = $8900 kg/m^3$

now we have

$m = (\frac{4}{3}\pi r^3)(8900)$

$m = (\frac{4}{3}\pi(\frac{0.007}{2})^3)(8900)$

$m = 1.6 gram$

now we have

$N_e = 9.63 \times 10^{23}$

now the charge on the copper ball is 40 nC

so the number of electrons removed

$Q = ne$

$40 \times 10^{-9} = n(1.6 \times 10^{-19}$

$n = 2.5 \times 10^{11}$

so the fraction of number of electrons removed is given as

$f = \frac{n}{N_e}$

$f = \frac{2.5 \times 10^{11}}{9.63 \times 10^{23}}$

$f = 2.6 \times 10^{-13}$

7 0

4 years ago