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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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When a researcher wants to correlate tension with depression, she develops tests that measure these two emotional states. The te
Igoryamba

Answer:

The answer is: measuring instrument

Explanation:

The researcher develops test to measure the variables she is interested in. So, they are her measuring instrument.

8 0
3 years ago
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What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s ​2 ​​ ) towards the Eart
Wittaler [7]

Answer:

The centripetal acceleration as a multiple of g=9.8 m/s^{2} is 1.020x10^{-3}m/s^{2}

Explanation:

The centripetal acceleration is defined as:

a = \frac{v^{2}}{r}  (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

v = \frac{2 \pi r}{T}  (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude 71.9^{\circ}. Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

\cos \theta = \frac{adjacent}{hypotenuse}

\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}} (3)

Where r_{71.9^{\circ}} is the radius at the latitude of 71.9^{\circ} and r_{e} is the radius at the equator (6.37x10^{6}m).

r_{71.9^{\circ}} can be isolated from equation 3:

r_{71.9^{\circ}} = r_{e} \cos \theta  (4)

r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})

r_{71.9^{\circ}} = 1.97x10^{6} m

Then, equation 2 can be used

v = \frac{2 \pi (1.97x10^{6} m)}{24h}

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

T = 24h . \frac{3600s}{1h} ⇒ 84600s

v = \frac{2 \pi (1.97x10^{6} m)}{84600s}

v = 146.31m/s

Finally, equation 1 can be used:

a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}

a = 0.010m/s^{2}

Hence, the centripetal acceleration is 0.010m/s^{2}

To given the centripetal acceleration as a multiple of g=9.8 m/s^{2}​ it is gotten:

\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}

6 0
3 years ago
A small sandbag is dropped from rest from a hovering hot-air balloon. (assume the positive direction is upward.) (a) after 1.5 s
marishachu [46]

solution:

We know v0 = 0, a = 9.8, t = 4.0. We need to solve for v

so,

we use the equation:

v = v0 + at

v = 0 + 9.8*4.0

v = 39.2 m/s

Now we just need to solve for d, so we use the equation:

d = v0t + 1/2*a*t^2

d = 0*4.0 + 1/2*9.8*4.0^2

d = 78.4 m

3 0
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7) the observation deck of the Empire State Building is 381 m above the street. Determine the
Elan Coil [88]

Answer:

Questions involving accelerating objects, including those in free fall, can be solved using three equations

 

y =  1/2at2 + v0t + y0

 

v = v0 + at

 

vf2 = v02 + 2 a (y - y0)

 

In this case, only the first equation is needed.

 

Since the object is falling the final height is 0 meters, the initial height is 370 m, and the acceleration is -9.8ms-2

 

Since the object is dropped, the initial velocity is 0.

 

0 = 1/2(-9.8)t2 + 370.    t = sqrt(370/4.9) = 8.69 s

Explanation:

8 0
4 years ago
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Suppose you wish to whirl a pail full of water in a vertical circle at a constant speed without spilling any of its contents (ev
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Answer:

V = 2.87 m/s

Explanation:

The minimum speed required would be that at which the acceleration due to gravity is negated by the centrifugal force on the water.

Thus, we simply need to set the centripetal acceleration equal to gravity and solve for the speed V using the following equation:

Centripetal acceleration = V^2 / r

where r is the distance of water from the pivot or shoulder.

For our case, r will be 0.65 + 0.19 = 0.84 m

and solving the above equation we get:

9.81 = V^2 / 0.84

V^2 = 8.2404

V = 2.87 m/s

6 0
3 years ago
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