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Katyanochek1 [597]
3 years ago
14

The maximum permissible workday dose for occupational exposure to radiation is 18 mrem. A 54-kg laboratory technician absorbs 2.

6 mJ of 0.30-MeV gamma rays in a workday. The relative biological effectiveness (RBE) for gamma rays is 1.00. What is the ratio of the equivalent dosage received by the technician to the maximum permissible equivalent dosage?
Physics
1 answer:
motikmotik3 years ago
6 0

Answer:

0.267 × 10⁻³

Explanation:

Given:

maximum permissible workday dose  = 18 mrem

Weight of the laboratory technician = 54 kg

gamma rays absorbed = 2.6 mJ of 0.30 MeV

relative biological effectiveness (RBE) for gamma rays = 1.00

Now,

Radiation Absorbed Dose = \frac{2.6 mJ}{54 kg}\times\frac{1J}{1\times10^3J}\times\frac{1 rad}{0.01 J/kg} = 4.8\times10^{-3}rad

also,

Roentgen Equivalent Man = rad × Q

here Q is the quality factor = 1 for gamma rays

thus,

Roentgen Equivalent Man = 4.8 × 10⁻³ × 1 = 4.8 × 10⁻³ rem

Therefore,

the ratio of the equivalent dosage received by the technician to the maximum permissible equivalent dosage will be

= \frac{4.8\times10^{-3}}{18}

or

= 0.267 × 10⁻³

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3 years ago
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A 5592 N piano is to be pushed up a(n) 3.79 m frictionless plank that makes an angle of 30.1 ◦ with the horizontal. Calculate th
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10628.87 J

Explanation:

We are given that

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3 years ago
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Train cars are coupled together by being bumped into one another. Suppose two loaded cars are moving toward one another, the fir
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Answer:

7560 Joules

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u_1 = Initial Velocity of first car = 0.3 m/s

u_2 = Initial Velocity of second car = -0.12 m/s

v = Velocity of combined mass

As linear momentum of the system is conserved

m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{1.5\times 10^5\times 0.3 + 2\times 10^5\times -0.12}{1.5\times 10^5 + 2\times 10^5}\\\Rightarrow v=0.06\ m/s

Energy lost is

\Delta E=\Delta E_i-\Delta E_f\\\Rightarrow \Delta=\frac{1}{2}(m_1u_1^2 + m_2u_2^2-(m_1+m_2)v^2)\\\Rightarrow \Delta=\frac{1}{2}(1.5\times 10^5\times 0.3^2 + 2\times 10^5\times (-0.12)^2-(1.5\times 10^5 + 2\times 10^5)\times 0.06^2)\\\Rightarrow \Delta=7560\ J

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3 years ago
A 4-kilogram ball moving at 8 m/sec to the right collides with a 1-kilogram ball at rest. After the collision,
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Conservation of momentum: total momentum before = total momentum after

Momentum = mass x velocity

So before the collision:
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32+0=32

Therefore after the collision
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What is the acceleration of a 0.30-kg volleyball when a player uses a force of 42 N to spike the ball?
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Answer:

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Then  acceleration a = F/m

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