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Sergio [31]
3 years ago
14

When cs-137 decays, it emits gamma radiation. the energy of one photon is 1.06 × 10-13 j. what is the wavelength of this radiati

on?
Physics
1 answer:
Kryger [21]3 years ago
3 0
To find the solution to the problem, we would be using Planck's equation which is E = hv
Where:
E = energy
h = Planck's constant = 6.626 x 10-34 J·s
ν = frequency
Then, you’ll need a second equation which is c = λν
Where:
c = speed of light = 3 x 108 m/sec
λ = wavelength
ν = frequency
Reorder the equation to solve for frequency:ν = c/λ
Next, substitute frequency in the first equation with c/λ to get a formula you can use:
E = hν
E = hc/λ
But we are looking for the wavelength, so rearrange it more, then our final equation would be:
λ = hc / E 
λ = (6.625E-34)(3.0E8 m/s) / (1.06E-13) 
λ = 1.875E-12 m
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A proton is moving in a circular orbit of radius 12 cm in a uniform 0.31-T magnetic field perpendicular to the velocity of the p
PIT_PIT [208]

Answer:

Explanation:

A proton of charge

q=+1.609×10^-19C

Orbit a radius of 12cm

r=0.12m

Magnetic Field of 0.31T

Angle between velocity and field is 90°

a. Because the magnetic force F supplies the centripetal force Fc.

The magnitude of the magnetic force F on a charge q moving at a speed v in a magnetic field of strength B is given by

F = qvB sin θ

And the centripetal force is given as

Fc=mv²/r

Where m is mass of proton

m=1.673×10^-27kg

Then, F=Fc

qvB sin θ=mv²/r

qBSin90=mv/r

rqB=mv

Then, v=rqB/m

v=0.12×1.609×10^-19×0.31/1.673×10^-23

v=3577692.78m/s

v=3.58×10^6m/s

b. Since,

F=qVBSin90

F=1.609×10^-19×3.58×10^6×0.31

F=1.785×10^-13 N.

6 0
3 years ago
The seafloor spreading process at ridges produces what kind of faults?
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4 0
3 years ago
Two wires A and B with circular cross-section are made of the same metal and have equal lengths, but the resistance of wire A is
Nesterboy [21]

Answer:

r₁/r₂ = 1/2 = 0.5

Explanation:

The resistance of a wire is given by the following formula:

R = ρL/A

where,

R = Resistance of wire

ρ = resistivity of the material of wire

L = Length of wire

A = Cross-sectional area of wire = πr²

r = radius of wire

Therefore,

R = ρL/πr²

<u>FOR WIRE A</u>:

R₁ = ρ₁L₁/πr₁²   -------- equation 1

<u>FOR WIRE B</u>:

R₂ = ρ₂L₂/πr₂²   -------- equation 2

It is given that resistance of wire A is four times greater than the resistance of wire B.

R₁ = 4 R₂

using values from equation 1 and equation 2:

ρ₁L₁/πr₁² = 4ρ₂L₂/πr₂²

since, the material and length of both wires are same.

ρ₁ = ρ₂ = ρ

L₁ = L₂ = L

Therefore,

ρL/πr₁² = 4ρL/πr₂²

1/r₁² = 4/r₂²

r₁²/r₂² = 1/4

taking square root on both sides:

<u>r₁/r₂ = 1/2 = 0.5</u>

6 0
3 years ago
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