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Marta_Voda [28]
3 years ago
7

A photon of wavelength 3.7 nm Compton scatters from an electron at an angle of 90°. What is the modified wavelength? (Enter your

answer correct to 5 significant figures.)

Physics
2 answers:
viva [34]3 years ago
7 0

Answer:

3.70242 nm

Explanation:

Using Compton effect formula

Δλ = ( h / mec) ( 1 - cosθ)

where h is planck constant = 6.62607 × 10 ⁻³⁴ m²kg/s

me, mass of an electron = 9.11 × 10⁻³¹ kg

c is the speed of light = 3 × 10⁸ m/s

Δλ = 6.62607 × 10 ⁻³⁴ m²kg/s / (9.11 × 10⁻³¹ kg × 3 × 10⁸ m/s ) ( 1 - cos 90°) =   0.242 × 10 ⁻¹¹ m = 2.42 × 10⁻¹² m  = 0.00242 nm

modified wavelength = 3.7 nm + 0.00242 nm = 3.70242 nm

denis-greek [22]3 years ago
7 0

Answer: 3.7024nm

Explanation:

in the attachment

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The statements below are all true. Some of them represent important reasons why the giant impact hypothesis for the Moon’s forma
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Answer:

the order of importance must be     b e a f c

Explanation:

Modern theories indicate that the moon was formed by the collision of a bad plant with the Earth during its initial cooling period, due to which part of the earth's material was volatilized and as a ring of remains that eventually consolidated in Moon.

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5 0
3 years ago
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.83 times a second. A tack is stuck in the tire a
timama [110]

Answer:

The tangential speed of the tack is 6.988 meters per second.

Explanation:

The tangential speed experimented by the tack (v), measured in meters per second, is equal to the product of the angular speed of the wheel (\omega), measured in radians per second, and the distance of the tack respect to the rotation axis (R), measured in meters, length that coincides with the radius of the tire. First, we convert the angular speed of the wheel from revolutions per second to radians per second:

\omega = 2.83\,\frac{rev}{s} \times \frac{2\pi\,rad}{1\,rev}

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7 0
2 years ago
A particle with charge −− 5.90 nCnC is moving in a uniform magnetic field B⃗ =−(B→=−( 1.28 TT )k^k^. The magnetic force on the p
maks197457 [2]

Answer:

Explanation:

Given that,

Charge q=-5.90nC

Magnetic field B= -1.28T k

And the magnetic force

F =−( 3.70×10−7N )i+( 7.60×10−7N )j

Let the velocity be V(xi + yj + zk)

Then, the force is given as

Note i×i=j×j×k×k=0

i×j=k. j×i=-k

j×k=i. k×j=-i

k×i=j. i×k=-j

The force in a magnetic field is given as

F= q(v×B)

−( 3.70×10−7N )i+( 7.60×10−7N )j =

q(xi + yj + zk) × -1.28k

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( -1.28x i×k - 1.28y j×k - 1.28z k×k)

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( 1.28xj - 1.28y i )

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( -1.28y i + 1.28x j)

So comparing comparing coefficients

let compare x axis component

-( 3.70×10−7N )i=-1.28qy i

−3.70×10−7N = -1.28qy

y= -3.7×10^-7/-1.28q

y= -3.7×10^-7/-1.28×-5.90×10^-9)

y=-48.99m/s

y≈-49m/s

Let compare y-axisaxis

7.6×10−7N j = 1.28qx j

7.6×10−7N = 1.28qx

x= 7.6×10^-7/1.28q

x= 7.6×10^-7/1.28×-5.90×10^-9

x=-100.64m/s

a. Then, the velocity of the x component is x= -100.64m/s

b. Also, the velocity component of the y axis is y=-49m/s

c. We will compute

V•F

V=-100.64i -49j

F=−( 3.70×10−7 N )i+( 7.60×10−7 N )j

Note

i.j=j.i=0. Also i.i = j.j =1

V•F is

(-100.64i-49j)•−(3.70×10−7N)i+(7.60×10−7 N )j =

3.724×10^-5 - 3.724×10^-5=0

V•F=0

d. Angle between V and F

V•F=|V||F|Cosx

0=|V||F|Cosx

Cosx=0

x= arccos(0)

x=90°

Since the dot product is zero, from vectors , if the dot product of two vectors is zero, then the vectors are perpendicular to each other

4 0
3 years ago
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