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barxatty [35]
2 years ago
12

An elementary particle of mass m completely absorbs a photon, after which its mass is 1.01m. (a) what was the energy of the inco

ming photon? (b) why is that energy greater than 0.01mc2?
Physics
1 answer:
sdas [7]2 years ago
8 0
A.) We use the famous equation proposed by Albert Einstein written below:

E = Δmc²
where
E is the energy of the photon
Δm is the mass defect, or the difference of the mass before and after the reaction
c is the speed of light equal to 3×10⁸ m/s

Substituting the value:

E = (1.01m - m)*(3×10⁸ m/s) = 0.01mc² = 3×10⁶ Joules

b) The actual energy may be even greater than 3×10⁶ Joules because some of the energy may have been dissipated. Not all of the energy will be absorbed by the photon. Some energy would be dissipated to the surroundings.
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The speed of light is constant in ALL frames of reference. That is 2.9 x10^8 m/s (290,000,000 m/s). It is independent of the motion of the light source.
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2 years ago
A bobsled is pushed with a force of 190.08 N. The sled has a mass of 28 kg. What is the acceleration of the bobsled? Report to t
Usimov [2.4K]
By definition it is known that force equals mass by acceleration. In other words F = m * a. To find the acceleration, you must clear the formula mentioned. Therefore, for a force of 190.08N and a mass of 28 Kg, we have that the acceleration is a = F / m = (190.08) / (28) = 6.79 m / s ^ 2
6 0
2 years ago
If in a vernier callipers 10 VSD coincides with 8 MSD then the least count of varnier calliper is
ratelena [41]

Answer:0.2mm

Explanation:

The length of one VSD=8/10=0.8mm

The least count of the instrument is the difference between the length of one MSD and length of one VSD

The length of inebriated MSD=1mm

Therefore,

The least count=1-0.8=0.2mm

3 0
2 years ago
Read 2 more answers
A woman fires a rifle with barrel length of 0.5400 m. Let (0, 0) be where the 125 g bullet begins to move, and the bullet travel
dybincka [34]

Answer:

Explanation:

Given that,

Length of barrel =0.54m

Mass of bullet=125g=0.125kg

Force extend

F=16,000+10,000x-26,000x²

a. Work done is given as

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=0.54m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=0.54m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=0.54m

W= 16,000(0.54) + 5000(0.54²) - 8666.667(0.54³) +0+0-0

W=8640+1458-1364.69

W=8733.31J

The workdone by the gas on the bullet is 8733.31J

b. Work done is given as

Work done when the length=1.05m

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=1.05m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=1.05m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=1.05mm

W= 16,000(1.05) + 5000(1.05²) - 8666.667(1.05³) +0+0-0

W=16800+5512.5-10032.75

W=12,279.75J

The workdone by the gas on the bullet is 12,279.75J

3 0
2 years ago
The frequency of the applied RF signal used to excite spins is directly proportional to the magnitude of the static magnetic fie
Anni [7]

Answer:

The inverse frequency is \dfrac{3}{80}\ s

Explanation:

Given that,

Magnetic field = 20 T

Proportionality constant = 5 Hz/T

Change in magnetic field = 3 T

We know that,

B=\dfrac{K}{\dfrac{1}{\omega}}

We need to calculate the inverse frequency

Using formula of frequency

\Delta(\dfrac{1}{\omega})=\dfrac{\Delta B}{k\times(\dfrac{1}{\omega^2})}

\Delta(\dfrac{1}{\omega})=\dfrac{k\times\Delta B}{B^2}

Put the value into the formula

\Delta(\dfrac{1}{\omega})=\dfrac{3\times5}{(20)^2}

\Delta(\dfrac{1}{\omega})=\dfrac{3}{80}\ s

Hence, The inverse frequency is \dfrac{3}{80}\ s

5 0
2 years ago
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