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grigory [225]
3 years ago
6

A 1-mCi source of^60 Co is placed in the center of a cylindrical water-filled tank with an inside diameter of 20 cm and depth of

100 cm. The tank is made of iron with a wall thickness of 1 cm. What is the uncollided flux density at the outer surface of the tank nearest the source?
Physics
1 answer:
LenKa [72]3 years ago
6 0

To solve this problem we need the concepts of Energy fluency and Intensity from chemical elements.

The energy fluency is given by the equation

\Psi=4RcE\pi

Where

\Psi =The energy fluency

c = Activity of the source

r = distance

E = electric field

In the other hand we have the equation for current in materials, which is given by

I= I_0 e^{-\mu_{h20}X_{h2o}} e^{-\mu_{Fe}X_{Fe}}

Then replacing our values we have that

I = 1*10^{-3} * 3.3*10^{10} * e ^{-0.06*1.1} e^{-0.058*7.861}

I = 1.3*10^7 Bq

We can conclude in this part that 1.3*10^7Bq is the activity coming out of the cylinder.

Now the energy fluency would be,

\Psi = \frac{cE}{4\pir^2}

\Psi = \frac{1.3*10^7*2*1.25}{4\pi*11^2}

\Psi = 2.14*10^4 MeV/cm^2.s

The  uncollided flux density at the outer surface of the tank nearest the source is \Psi = 2.14*10^4 MeV/cm^2.s

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Two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision
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a) The velocity after the collision.is 11.456 m/s.

b) The kinetic energy lost due to the collision is 44.564 J.

<h3>What is conservation of momentum principle?</h3>

When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.

The external force is not acting here, so the initial momentum is equal to the final momentum. For inelastic collision, final velocity is the common velocity for both the bodies.

m₁u₁ +m₂u₂ =(m₁ +m₂) v

Given are the two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision. Their initial velocities along the one-dimension path are vi1 = 32.4 m/s [right] and vi2 = 8.6 m/s [left].

(a) Substitute the values, then the final velocity will be

0.6 x32.4 +4.4 x 8.6 = (0.6+4.4)v

v = 11.456 m/s

Thus, the velocity after collision is 11.456 m/s.

(b) Kinetic energy lost due to collision will be the difference between the kinetic energy before and after collision.

= [1/2m₁u₁² +1/2m₂u₂² ] - [1/2(m₁ +m₂) v²]

Substitute the value, we have

= [1/2 x 0.6 x32.4² + 1/2 x4.4 x 8.6²] - [1/2 x(0.6+4.4)11.456²]

= 44.564 J

Thus, the kinetic energy lost due to the collision is 44.564 J.

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