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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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Two long, parallel wires separated by 2.00 cm carry currents in opposite directions. The current in one wire is 1.75 A, and the
Natasha_Volkova [10]

The force per unit length between the two wires is 6.0\cdot 10^{-5} N/m

Explanation:

The magnitude of the force per unit length exerted between two current-carrying wires is given by

\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}

where

\mu_0 = 4\pi \cdot 10^{-7} Tm/A is the vacuum permeability

I_1, I_2 are the currents in the two wires

r is the separation between the two wires

For the wires in this problem, we have

I_1 = 1.75 A

I_2 = 3.45 A

r = 2.00 cm = 0.02 m

Substituting into the equation, we find

\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(1.75)(3.45)}{2\pi (0.02)}=6.0\cdot 10^{-5} N/m

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4 0
3 years ago
14. a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s. a. how much later does the ball
Burka [1]

The time taken to hit the ground is 3.9 s, the range is 18m and the final velocity is 42.82 m/s

<h3>Motion Under Gravity</h3>

The motion of an object under gravity is the vertical motion of the object under the influence of acceleration due to gravity.

Given that a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s.

a. how much later does the ball hit the ground?

The time can be calculated by considering the vertical component of the motion with the use of formula below.

h = ut + 1/2gt²

Where

  • Height h = 75 m
  • Initial velocity u = 0 ( vertical velocity )
  • Acceleration due to gravity g = 9.8 m/s²
  • Time t = ?

Substitute all the parameters into the formula

75 = 0 + 1/2 × 9.8 × t²

75 = 4.9t²

t² = 75/4.9

t² = 15.30

t = √15.3

t = 3.9 s

b. how far from the building will it land?

The range can be found by using the formula

R = ut

Where u = 4.6 m/s ( horizontal velocity )

R = 4.6 × 3.9

R = 18 m

c. what is the velocity of the ball just before it hits the ground?

The final velocity will be

v = u + gt

v = 4.6 + 9.8 × 3.9

v = 4.6 + 38.22

v = 42.82 m/s

Therefore, the answers are 3.9 s, 18 m and 42.82 m/s

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7 0
1 year ago
A pendulum is observed to complete 20 full cycles in 60 seconds. Determine the period and the frequency of the pendulum.
Delvig [45]

i hope i have been useful buddy.

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5 0
3 years ago
Read 2 more answers
A 43.9-g piece of copper (CCu= 0.385 J/g°C) at 135.0°C is plunged into 254 g of water at 39.0°C. Assuming that no heat is lost t
Semmy [17]

Answer:

T = 40.501\,^{\textdegree}C

Explanation:

The interaction of the piece of copper and water means that the first one need to transfer heat in order to reach a thermal equilibrium with water. Then:

-Q_{out,Cu} = Q_{in,H_{2}O}

After a quick substitution, the expanded expression is:

-(43.9\,g)\cdot (0.385\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-135^{\textdegree}C) = (254\,g)\cdot (4.187\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-39\,^{\textdegree}C)

-16.902\,\frac{J}{^{\textdegree}C}\cdot (T-135^{\textdegree}C) = 1063.498\,\frac{J}{^{\textdegree}C} \cdot  (T-39^{\textdegree}C)

43758,192\,J = 1080.4\,\frac{J}{^{\textdegree}C}\cdot T

The final temperature of the system is:

T = 40.501\,^{\textdegree}C

8 0
3 years ago
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How does the atmosphere’s interaction with different wavelengths of light help to protect life on earth?
marishachu [46]

Answer:

Ozone layer in the upper atmosphere filters most of the harmful radiations of shorter wavelength. It actually absorbs the hazardous radiations like ultraviolet, gamma rays, x- rays and most of all those having shorter wavelength then the visible light. That's how the earth's atmosphere protects life on earth. But unfortunately, climate change and global warming is causing the depletion of ozone layer which is causing skin related diseases and harming not only the human life but also the plants and animals.

4 0
3 years ago
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