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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 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.
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(a) 69.3 J

The work done by the applied force is given by:

W=Fd cos \theta

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

\theta=30^{\circ} is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

\Delta K = W

where

\Delta K is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

\Delta K=69.3 J

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, \Delta K, after it has travelled for d=3 m. Using the work-energy theorem again, we find

\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J

And substituting into (1) and re-arrangin the equation, we find

v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s

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