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n200080 [17]
4 years ago
7

An x-ray photon is scattered by an originally stationary electron. how does the frequency of the scattered photon compare relati

ve to the frequency of the incident photon? it is lower.
Physics
1 answer:
Viefleur [7K]4 years ago
8 0

The frequency of the scattered photon decreases or it will be lower compare to the frequency of incident photon. An x-ray photon scatters in one direction after a collision and some energy is transferred to the electron as it recoils in another direction resulting to have less energy in the scattered photon. In addition, the frequencies will also depend on the differences of the angle at which the scattered photon leaves the collision and this incident is called Compton Effect.

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ivanzaharov [21]
The answer would be 46.482 because you multiply 18.3 by 2.54 because for every inch you get 2.54 centimeters
6 0
2 years ago
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The intensity level of a "Super-Silent" power lawn mower at a distance of 1.0 m is 100 dB. You wake up one morning to find that
OLEGan [10]

Answer: The intensity level of sound in the bedroom is 80dB

Explanation:

Intensity of lawn mower at r=1m is 100dB

Beta1= 10dBlog(I1/Io)

100dB= 10dB log(I1/Io)

10^10= I1/Io

I1= Io(10^10)

10^12)×(10^10)= I1

I1=10^-2w/m^2

Intensity of lawn mower at r=20m

I2/I1=(r1/r2)^2 =(1/20)^2

I2= I1(1/400)

I2=2.5×10^-3W_m^2

Intensity of 4 lown mowers at 20m fro. Window

= 10dBlog(4I2/Io)

= 10^-4/10^-12

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6 0
3 years ago
A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit
Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M=8.7\cdot 10^{13}kg is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

m\frac{v^2}{r}=G\frac{Mm}{r^2}

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s

3 0
4 years ago
An amplifier has a 50 watt output and a 5 watt input. what is the gain in decibels for this amplifier, rounded to the nearest de
Marrrta [24]
Gain in decibels is given by;

Gain db = 10*log (Po/Pi), where Po = Power output, Pin = Power input

Substituting;

Gain in db = 10 * log (50/5) = 10 db
6 0
3 years ago
A circular loop of wire with radius r=0.0250 m and resistance r=0.390 ohms is in a region of spatially uniform magnetic field. t
slavikrds [6]

Answer:

0.0133 A

Explanation:

The time at which B=1.33 T is given by  

1.33 = 0.38*t^3  

t = (1.33/0.38)^(1/3) = 1.52 s  

Using Faraday's Law, we have  

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Area A = pi * r² = 3.141 *(0.025 *0.025) = 0.00196 m²

emf = - A*(3*0.38)*t^2  

thus, the emf at t=1.52 s is  

emf = - 0.00196*(3*0.38)*(1.52)^2 = -0.0052 V  

if the resistance is 0.390 ohms, then the current is given by  

I = V/R = 0.0052/0.390 = 0.0133 A

3 0
3 years ago
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