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anzhelika [568]
3 years ago
6

a 10-N force is exerted on a box, moving it 20 m in the same direction. What is the magnitude of work done on the box?

Physics
1 answer:
-Dominant- [34]3 years ago
8 0
F = 10 N

d = 20 m

θ = 0°

W = F (dot product) D = F * D * cos(angle between them)

W = FDcosθ

W = 10 * 20 * cos0 = 200 J
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For crystal diffraction experiments, wavelengths on the order of 0.25 nm are often appropriate.
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A) E = 4.96 x 10³ eV

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Explanation:

A)

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E = energy of photon = ?

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<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
3 years ago
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