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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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Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m
Phoenix [80]

Answer:

a. t_1=12.5\ s

b. a_2=-13.61\ m.s^{-2}  must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. t_2=2.5714\ s is the time taken to stop after braking

Explanation:

Given:

  • speed of leading car, u_1=25\ m.s^{-1}
  • speed of lagging car, u_{2}=35\ m.s^{-1}
  • distance between the cars, \Delta s=45\ m
  • deceleration of the leading car after braking, a_1=-2\ m.s^{-2}

a.

Time taken by the car to stop:

v_1=u_1+a_1.t_1

where:

v_1=0 , final velocity after braking

t_1= time taken

0=25-2\times t_1

t_1=12.5\ s

b.

using the eq. of motion for the given condition:

v_2^2=u_2^2+2.a_2.\Delta s

where:

v_2= final velocity of the chasing car after braking = 0

a_2= acceleration of the chasing car after braking

0^2=35^2+2\times a_2\times 45

a_2=-13.61\ m.s^{-2} must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

v_2=u_2+a_2.t_2

0=35-13.61\times t_2

t_2=2.5714\ s  is the time taken to stop after braking

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3 years ago
A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the mer
son4ous [18]

Solution :

Given data :

Mass of the merry-go-round, m= 1640 kg

Radius of the merry-go-round, r = 7.50 m

Angular speed, $\omega = \frac{1}{8}$  rev/sec

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                              = 5.89 rad/sec

Therefore, force required,

$F=m.\omega^2.r$

   $$=1640 \times (5.89)^2 \times 7.5  

   = 427126.9 N

Thus, the net work done for the acceleration is given by :

W = F x r

   = 427126.9 x 7.5

   = 3,203,451.75 J

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Answer:

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One this evidence can be seen from the operation of pin-hole camera, which depends on rectilinear propagation of light

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