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3 years ago

How is the frequency of an electromagnetic wave related to energy

Physics

1 answer:

aivan3 [116]3 years ago

7 0

Answer: The energy carried by an electromagnetic wave is proportional to the frequency of the wave.

Explanation:

Electromagnetic waves that are of higher energy than visible light (higher frequency, shorter wavelength) include ultraviolet light, X-rays, and gamma rays.

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Two cars, C and D, travel in the same direction on a long, straight section of highway. During a particular time interval Ato, c

a_sh-v [17]

Answer:

possibly because the car is running out of gas

Explanation:

6 0

3 years ago

Electrical power companies sell electrical energy

mixer [17]

Heat used by electric heater :

Q = m • c • ∆T

Q = (75 kg)(4200 J/kg°C)(43°C - 15°C)

Q = 8.82 × 10⁶ J

Cost of electrical energy :

Cost = (8.82 × 10⁶ J)/(3.6 × 10⁶ J) • ($ 0.15)

Cost = $ 0.3675

4 0

2 years ago

An object weighs 32 newtons. What is its mass if a gravitometer indicates that g = 8.25 m/s?

Elan Coil [88]

Explanation:

weight=mass×g

32=mass×8.25

mass=

$\frac{128}{33} = 3.878$ kg

8 0

2 years ago

At the Indianapolis 500, you can measure the speed of cars just by listening to the difference in pitch of the engine noise betw

allsm [11]

To develop this problem it is necessary to apply the concepts related to the Dopler effect.

The equation is defined by

$f_i = f_0 \frac{c}{c+v}$

Where

$f_h$ = Approaching velocities

$f_i$ = Receding velocities

c = Speed of sound

v = Emitter speed

And

$f_h = f_0 \frac{c}{c+v}$

Therefore using the values given we can find the velocity through,

$\frac{f_h}{f_0}=\frac{c-v}{c+v}$

$v = c(\frac{f_h-f_i}{f_h+f_i})$

Assuming the ratio above, we can use any f_h and f_i with the ratio 2.4 to 1

$v = 353(\frac{2.4-1}{2.4+1})$

$v = 145.35m/s$

Therefore the cars goes to 145.3m/s

7 0

3 years ago

If an X-ray beam of wavelength 1.4 × 10-10 m makes an angle of 20° with a set of planes in a crystal causing first order constru

Flura [38]

Answer:

43.16°

Explanation:

λ = Wavelength = 1.4×10⁻¹⁰ m

θ₁ = 20°

n can be any integer

d = distance between the two slits

Since for the first bright fringe, n₁ = 1

n₂ = 2 for second order line

The relation between the distance of the slits and the angle through which it is passed is:

dsinθ=nλ

As d and λ are constant

$\frac{n_1\lambda}{sin \theta_1}=\frac{n_2\lambda}{sin \theta_2}\\\Rightarrow \frac{1}{sin20}=\frac{2}{sin\theta_2}\\\Rightarrow sin\theta_2=\frac{2}{\frac{1}{sin20}}\\\Rightarrow \theta_2=sin^{-1}{\frac{2}{\frac{1}{sin20}}}\\\Rightarrow \theta_2=43.16^{\circ}$

∴ Angle by which the second order line appear is 43.16°

5 0

3 years ago