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ahrayia [7]
3 years ago
6

A photon with wavelength λ = 0.0830 nm is incident on an electron that is initially at rest. if the photon scatters in the backw

ard direction, what is the magnitude of the linear momentum of the electron just after the collision with the photon?
Physics
1 answer:
QveST [7]3 years ago
7 0
Compton scattering equation of wavelengths 
λ'-λ = h/mec (1 - CosФ)
λ' = λ + h/mec (1 - cos 180°)
= ( 0.0830nm) + (6.626 × 10⁻³⁴ J.s)/ (9.1 × 10⁻³¹ kg)(3.0 × 10⁸ m/s
= 0.0830nm
The momentum of electron is
P photon λ = Pe + P phpton λ'
Pe = h/λ - ( -h/λ') = h(λ' + λ)/λλ'
= (6.626 × 10⁻³⁴ J.s)(( 0.08785nm) +( 0.0883nm)/0.08785nm)( 0.083nm)
1.55 × 10⁻²³ kg.m/s.
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Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote
Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of 1.8 \times 10^{-8}\; \rm C each, the electric potential at the midpoint of segment \rm AB would be approximately 8.3 \times 10^{3}\; \rm V.

Explanation:

Convert the unit of the length of each side of this triangle to meters: 10\; \rm cm = 0.10\; \rm m.

Distance between the midpoint of \rm AB and each of the three charges:

  • d({\rm A}) = 0.050\; \rm m.
  • d({\rm B}) = 0.050\; \rm m.
  • d({\rm C}) = \sqrt{3} \times (0.050\; \rm m).

Let k denote Coulomb's constant (k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}.)

Electric potential due to the charge at \rm A: \displaystyle \frac{k\, q}{d({\rm A})}.

Electric potential due to the charge at \rm B: \displaystyle \frac{k\, q}{d({\rm B})}.

Electric potential due to the charge at \rm A: \displaystyle \frac{k\, q}{d({\rm C})}.

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of \rm AB due to all these three charges  would be:

\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}.

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an 269 kg object is moved a distance of 1.9 m by a force if 580 j of work is done on the object what is the object acceleration
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7 0
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A backyard swimming pool with a circular base of diameter 6.00m is filled to depth 1.50m. (b) Two persons with combined mass 150
Mashutka [201]

The pressure increase at the bottom of the pool after they enter the pool and float is 106.103 Pa.

<h3>What is absolute pressure?</h3>

Absolute pressure is the force that exists in a space when there is no matter present, or when there is a perfect vacuum. This absolute zero serves as the baseline for measurements in absolute pressure. The measurement of barometric pressure is the greatest illustration of an absolute referenced pressure. In order to determine absolute pressure, a complete vacuum is used. In contrast, gauge pressure is the amount of pressure that is measured in relation to atmospheric pressure, also referred to as barometric pressure.

given,

diameter = 6 m

depth = h = 1.5 m

Atmospheric pressure = P₀ = 10⁵ Pa

a) absolute pressure

P = P₀ + ρ g h

P = 10⁵ + 1000 x 10 x 1.5

P = 1.15 x 10⁵ Pa

b) When two person enters into the pool,

mass of the two person = 150 Kg

weight of water level displaced exists equal to the weight of person.

\rho \mathrm{Vg}=2 \mathrm{mg} \\

V=\frac{2 m}{\rho} \\

V=\frac{2 \times 150}{1000} \\

\mathrm{~V}=0.3 \mathrm{~m}^3

Area of pool $=\frac{\pi}{4} d^2$

&=\frac{\pi}{4} \times 6^2 \\

&=28.27 \mathrm{~m}^2

Height of the water rise

h &=\frac{V}{A} \\

h &=\frac{0.3}{28.27} \\

& \mathrm{~h}=0.0106 \mathrm{~m}

  • Pressure increased

P = ρ g h

P = 1000 x 10 x 0.0106

P = 106.103 Pa

To learn more about absolute pressure refer to:

brainly.com/question/17200230

#SPJ4

4 0
1 year ago
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