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Bas_tet [7]
3 years ago
12

Gamma-ray bursters are objects in the universe that emit pulses of gamma rays with high energies. The frequency of the most ener

getic bursts has been measured at around 3.0 × 1021 Hz. The speed of light is 3 × 108 m/s. What is the wavelength of these gamma rays? Answer in units of m
Physics
1 answer:
Zielflug [23.3K]3 years ago
4 0
Wavelength and frequency are related and they are indirectly related The higher the frequency of a light, the shorter the wavelength would be. The constant of proportionality of this relation is the speed of light. We calculate as follows:
wavelength = speed of light / frequency
wavelength = 3x10^8 / 3.0x10^21
wavelength = 1x10^-13 m
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A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 6 ft from the path and is kept fo
BARSIC [14]

We are given that,

\frac{dx}{dt} = 4ft/s

We need to find \frac{d\theta}{dt} when x=8ft

The equation that relates x and \theta can be written as,

\frac{x}{6} tan\theta

x = 6tan\theta

Differentiating each side with respect to t, we get,

\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}

\frac{dx}{dt} = (6sec^2\theta)\cdot \frac{d\theta}{dt}

\frac{d\theta}{dt} = \frac{1}{6sec^2\theta} \cdot \frac{dx}{dt}

Replacing the value of the velocity

\frac{d\theta}{dt} = \frac{1}{6} cos^2\theta (4)^2

\frac{d\theta}{dt} = \frac{8}{3} cos^2\theta

The value of cos \theta could be found if we know the length of the beam. With this value the equation can be approximated to the relationship between the sides of the triangle that is being formed in order to obtain the numerical value. If this relation is known for the value of x = 6ft, the mathematical relation is obtained. I will add a numerical example (although the answer would end in the previous point) If the length of the beam was 10, then we would have to

cos\theta = \frac{6}{10}

\frac{d\theta}{dt} = \frac{8}{3} (\frac{6}{10})^2

\frac{d\theta}{dt} = \frac{24}{25}

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4 0
3 years ago
How are electromagnetic waves that are produced by oscillating charges and sound waves that are produced by oscillating tuning f
Dovator [93]
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8 0
4 years ago
Glass is transparent to visibile light under normal conditions; however, at extremely high intensities, glass will absorb most o
8_murik_8 [283]

Answer:

3 photons

Explanation:

The energy of a photon E can be calculated using this formula:

E=\frac{hc}{\lambda}

Where h corresponds to Plank constant (6.626070x10^-34Js), c is the speed of light in the vacuum (299792458m/s) and \lambda is the wavelength of the photon(in this case 800nm).

E=\frac{hc}{\lambda}=\frac{(6.626070\times10^{-34})(299792458)}{800\times10^{-9}}=\frac{1.986445812\times10^-25}{800}=2.483057265\times10^{-19}J

Tranform the units

1eV=1.602176634\times10^{-19}J\\2.483057265\times10^{-19}J(\frac{1eV}{1.602176634\times10^{-19}J})=1.549802445eV

The band Gap is 4eV, divide the band gap between the energy of the photon:

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Rounding to the next integrer: 3.

Three photons are the minimum to equal or exceed the band gap.

4 0
3 years ago
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