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3 months ago

6

Gamma rays may be used to kill pathogens in ground beef. One irradiation facility uses a 60Co source that has an activity of 1.0

×106Ci. 60Co undergoes beta decay and then gives off two gamma rays, at 1.17 and 1.33 MeV; typically 30% of this gamma-ray energy is absorbed by the meat. The dose required to kill all pathogens present in the beef is 4000 Gy.
How many kilograms of meat per hour can be processed in this facility? Express your answer in kilograms per hour.

1 answer:

ValentinkaMS [3.4K]3 months ago

4 0

Answer:

Explanation:

$C_i=3.7\times 10^{16} \,decays/sec$

The energy from gamma rays generated by ore decay is

$(1.17+1.33)MeV=2.50MeV$

The energy from gamma rays created per second is

$(2.5\times 3.7\times 10^{16})MeV$

Activity

$1 \times 10^6c$

The total energy of gamma rays produced in one second is

$(2.5 \times 3.7 \times 10^{16} \times 10^6)MeV$

E represents the energy of gamma rays generated in one hour is

$(2.5\times 3.7\times 10^{10}\times 60 \times 60)MeV$

Gamma ray energy absorbed by meat = 30% (in 1 hour) of E is 0.3E

The dosage necessary to eliminate pathogens = 4000Gy=4000J/kg

The mass of meat that can be processed every hour is

$\frac{0.3E/hr}{4000J/kg}\\\\E=(2.5\times 3.7\times 10^{10}\times 60 \times 60)MeV\\\\=333\times 10^{18}MeV\\\\eV=1.6\times 10^{-19}\\\\E=53.3J$

The total meat produced = $\frac{0.3\times 53.3}{4000}=3.996\times 10^{-3}$

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2 months ago

At time t=0 a proton is a distance of 0.360 m from a very large insulating sheet of charge and is moving parallel to the sheet w

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Explanation:

The formula for the electric field produced by an infinite sheet of charge is outlined below.

E = $\frac{\sigma}{2 \epsilon_{o}}$

where, $\sigma$ is the surface charge density

Following this, the formula for the electric force acting on a proton is given as:

F = eE

where, e is the charge of a proton

According to Newton's second law of motion, the overall force on the proton can be expressed as follows.

F = ma

a = $\frac{eE}{m}$

= $\frac{e(\frac{\sigma}{2 \epsilon_{o}})}{m}$

= $\frac{e \sigma}{2m \epsilon_{o}}$

According to kinematic equations, the proton's speed in the perpendicular direction can be described as follows.

$v_{f} = v_{i} + at$

= $(0 m/s) + \frac{e \sigma}{2 m \epsilon_{o}}t$

= $\frac{1.6 \times 10^{-19}C \times 2.34 \times 10^{-9} C/m^{2} \times 5.40 \times 10^{-8}s}{2 \times (1.67 \times 10^{-27} kg)(8.85 \times 10^{-12} C^{2}/Nm^{2}}$

= 683.974 m/s

Thus, the overall speed of the proton can be calculated as follows.

v' = $\sqrt{(960 m/s)^{2} + (683.974 m/s)^{2}}$

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= $\sqrt{1389420.43}$

= 1178.73 m/s

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2 months ago

An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:

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Heat supplied to the gas = Q = 743 Joules

Work applied to the gas = W = -743 Joules

$\texttt{ }$

Additional explanation

The Ideal Gas Law that should be remembered is:

$\large {\boxed {PV = nRT} }$

P = Pressure (Pa)

V = Volume (m³)

n = number of moles (moles)

R = Gas Constant (8.314 J/mol K)

T = Absolute Temperature (K)

Now, let’s proceed with the problem!

$\texttt{ }$

Given:

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

Unknown:

Work done on the gas = W =?

Heat supplied to the gas = Q =?

Solution:

Step A:

An ideal gas expands isothermally:

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

$V_2 = 10 \div 3$

$V_2 = 3\frac{1}{3} \texttt{ L}$

$\texttt{ }$

Next, we will determine the work performed on the gas:

$W_A = -P_2(V_2 - V_1)$

$W_A = -3.00(3\frac{1}{3} - 2.00)$

$W_A = \boxed{-4 \texttt{ L.atm}}$

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Step B:

By utilizing the methodology mentioned earlier:

$P_2V_2 = P_3V_3$

$3.00 \times 3\frac{1}{3} = 2.00 \times V_3$

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

$\texttt{ }$

Next, we will ascertain the work completed on the gas:

$W_B = -P_3(V_3 - V_2)$

$W_B = -2.00(5 - 3\frac{1}{3})$

$W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}$

$\texttt{ }$

Ultimately, we can calculate the total work done and heat supplied as follows:

$W = W_A + W_B$

$W = -4 + (-3\frac{1}{3})$

$W = -7\frac{1}{3} \texttt{ L.atm}$

$W = -7\frac{1}{3} \times 101.33 \texttt{ J}$

$\boxed{W \approx -743 \textt{ J}}$

$\texttt{ }$

$\Delta U = Q + W$

$0 = Q + (-743)$

$\boxed{Q = 743 \texttt{ J}}$

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Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

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2 months ago

Assume that the cart is free to roll without friction and that the coefficient of static friction between the block and the cart

Keith_Richards [3271]

Answer: $F=\frac{(M+m)g}{\mu _s}$

Explanation:

Provided:

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A force F is applied to mass m.

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The frictional force will counterbalance the weight of the block.

The frictional force is $=\mu _sN$

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$\mu _sN=mg$

$\mu _sma=mg$

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$F=\frac{(M+m)g}{\mu _s}$

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3 months ago

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