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adell [148]
3 years ago
13

What potential increase vac must an electron be accelerated through if the most energetic photon it can emit will scatter off of

a stationary electron at an angle ϕ=60∘ with wavelength 8.4×10−12m?express your answer in kilovolts to two significant digits?
Physics
1 answer:
nikitadnepr [17]3 years ago
7 0
Referring to Compton scattering 
Δλ = h/m₀c (I- cos Ф)
λ' =λ = (0,0242×10⁻¹⁰) (1- cos 60°)
λ= λ' -(0.0242 × 10⁻¹⁰) (1- cos 60°)

7.19 ˣ 10⁻¹²m

The increased potential is given by 
Vₐc = hc/eλ = 6.625 × 10 ⁻³⁴ J,s) ( 3× 10⁸ m/s ( 1.6 ˣ 10 ⁻¹⁰C)
(7.19 ˣ 10⁻¹²m)

173kV.
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Determine (a) the starting height, (b) the time to hit the ground, and (c) the velocity when it hits the ground for an object sh
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Answer:

a

Explanation:

8 0
3 years ago
ball A of mass 5.0 kg moving at 20 meters per second collides with ball B of unknown mass moving at 10. meters per second in the
antiseptic1488 [7]

Answer:The mass of ball B is 10 kg.

Explanation;

Mass of ball A = M_A=5 kg

Velocity of the ball A before collision:U_A=20 m/s

Velocity of ball A after collision=V_A=10 m/s

Mass of ball B= M_B

Velocity of the ball B before collision:U_B=10 m/s

Velocity of ball B after collision=V_B=15 m/s

M_AV_A+M_BV_B=M_AU_A+M_BU_B

5 kg\times 10 m/s+M_B\times 15=5 kg\times 20m/s+M_B\times 10m/s

M_B=10kg

The mass of ball B is 10 kg.

8 0
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harina [27]

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Energy

Explanation:

4 0
3 years ago
Three forces act on an object. Two of the forces are at an angle of 100◦to each other and have magnitude 25N and 12N. The third
seraphim [82]

Answer:

F₄ = 29.819 N

Explanation:

Given

F₁ = (- 25*Cos 50° i + 25*Sin 50° j + 0 k) N

F₂ = (12*Cos 50° i + 12*Sin 50° j + 0 k) N

F₃ = (0 i + 0 j + 4 k) N

Then we have

F₁ + F₂ + F₃ + F₄ = 0

⇒   F₄ = - (F₁ + F₂ + F₃)

⇒   F₄ = - ((- 25*Cos 50° i + 25*Sin 50° j) N + (12*Cos 50° i + 12*Sin 50° j) N + (4 k) N) = (13*Cos 50° i - 37*Sin 50° j - 4 k) N

The magnitude of the force will be

F₄ = √((13*Cos 50°)² + (- 37*Sin 50°)² + (- 4)²) N = 29.819 N

6 0
3 years ago
calculate the resistance of a wire 150cm long and diameter 2.0mm constructed from an alloy of resistivity 44*10-⁸Ωm​
Ghella [55]

Answer:

R = 0.21 Ω

Explanation:

the formula:

R = r x l/A

R = (44 x 10-⁸ Ωm) x 1.5 / (π x (1 x 10-³ m)²)

R = 6.6 x 10-⁷ / 3.14 x 10-⁶

R = 0.21 Ω

8 0
3 years ago
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