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3 years ago

11

Visible light travels at a speed 3.0 × 108 of m/s. If red light has a wavelength of 6.5 × 10–7 m, what the frequency of this lig

ht?
A) 2.2 × 10–15 HZ
B) 2.0 × 102 HZ
C) 3.0 × 108 HZ
D) 4.6 × 1014 HZ

2 answers:

WARRIOR [948]3 years ago

8 0

Answer on Edge is D.

Frequency = (3/6.5) x 10¹⁵ = 0.4615 x 10⁻¹⁵ = 4.615 x 10⁻14 second.

:)

gayaneshka [121]3 years ago

4 0

i think its d on edge

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An aircraft with a mass of 10,000 kg starts from rest at sea level and takes off, then flies to a cruising speed of 620 km/h and

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Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

$\Delta P.E=mg(h_{2}-h_{1})$

$\Delta P.E=10000\times9.8\times(10000-0)$

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For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

$\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)$

$\Delta K.E=148298642\ J$

$\Delta K.E=148.3\ MJ$

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

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Mass attached to the string, m = 0.200 kg

Distance, d = 30 cm = 0.03 m

Now,

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$Mgsin\theta + \mu_{k}Mgcos\theta + Ma = T$ (1)

Also

T = m(g + a)

Thus eqn (1) can be written as:

$Mgsin\theta + \mu_{k}Mgcos\theta + Ma = m(g - a)$

$Mgsin\theta + \mu_{k}Mgcos\theta + Ma = mg + ma$

$mg - Mgsin\theta - \mu_{k}Mgcos\theta = (M - m)a$

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3 0

2 years ago

(i) Calculate the energy, in electron volts, of a photon whose frequency is (a) 620 THz}, (b) 3.10GHz , and(c) 46.0 MHz

DedPeter [7]

The energy in electron volts of the photons that has the following frequencies is as follows:

620 THz = 2.564eV
3.10GHz = 1.28 × 10-⁵eV
46.0 MHz = 1.902 × 10-⁷eV

<h3>How to calculate energy?</h3>

The energy of a photon can be calculated using the following formula:

E = hf

Where;

E = energy
h = Planck's constant (6.626 × 10-³⁴ J/s)
f = frequency

First, we convert the frequencies to hertz as follows;

620THz = 6.2 × 10¹⁴Hz
3.10GHz = 3.1 × 10⁹Hz
46.0MHz = 4.6 × 10⁷Hz

E = 6.626 × 10-³⁴ × 6.2 × 10¹⁴ = 2.564eV
E = 6.626 × 10-³⁴ × 3.1 × 10⁹ = 1.28 × 10-⁵eV
E = 6.626 × 10-³⁴ × 4.6 × 10⁷ = 1.902 × 10-⁷eV

Learn more about energy of a photon at: brainly.com/question/2393994

#SPJ1

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1 year ago