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taurus [48]
2 years ago
7

what was the temperature change in Celsius degree if it is changed from 44 degree fahrenheit to -56 degree fahrenheit​

Physics
1 answer:
wel2 years ago
5 0

Answer:

-11.11 degree Celsius

Explanation:

The change was 44 degree fanhereit

To 56 degree fanhereit

Therefore the temperature range can be calculated as follows

56-44

= 12 degree fanhereit to Celsius

= 12-32×5/9

= -20×5/9

= 100/9

= -11.11 degree Celsius

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When an electron moves 2.5 m in the direction of an electric field, the change in electrical potential energy of the electron is
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7 0
3 years ago
For crystal diffraction experiments, wavelengths on the order of 0.25 nm are often appropriate.
Kamila [148]

Answer:

A) E = 4.96 x 10³ eV

B) E = 4.19 x 10⁴ eV

C) E = 3.73 x 10⁹ eV

Explanation:

A)

For photon energy is given as:

E = hv

E = \frac{hc}{\lambda}

where,

E = energy of photon = ?

h = 6.625 x 10⁻³⁴ J.s

λ = wavelength = 0.25 nm = 0.25 x 10⁻⁹ m

Therefore,

E = \frac{(6.625 x 10^{-34} J.s)(3 x 10^8 m/s)}{0.25 x 10^{-9} m}

E = (7.95 x 10^{-16} J)(\frac{1 eV}{1.6 x 10^{-19} J})

<u>E = 4.96 x 10³ eV</u>

<u></u>

B)

The energy of a particle at rest is given as:

E = m_{0}c^2

where,

E = Energy of electron = ?

m₀ = rest mass of electron = 9.1 x 10⁻³¹ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

E = (9.1 x 10^{-31} kg)(3  x  10^8 m/s)^2\\

E = (8.19 x 10^{-14} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\

<u>E = 4.19 x 10⁴ eV</u>

<u></u>

C)

The energy of a particle at rest is given as:

E = m_{0}c^2

where,

E = Energy of alpha particle = ?

m₀ = rest mass of alpha particle = 6.64 x 10⁻²⁷ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

E = (6.64 x 10^{-27} kg)(3  x  10^8 m/s)^2\\

E = (5.97 x 10^{-10} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\

<u>E = 3.73 x 10⁹ eV</u>

8 0
2 years ago
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