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Pani-rosa [81]
2 years ago
13

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 work day. 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?
a) .37
b) .35
c) .27
d) .29
e) .32
Physics
1 answer:
professor190 [17]2 years ago
6 0
32 bsbsbsnsbzhhzsbnsnsnsnsnsjs
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(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0
3 years ago
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