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GenaCL600 [577]
3 years ago
15

It has become popular for some people to have yearly whole-body scans (CT scans, formerly called CAT scans), using x-rays, just

to see if they detect anything suspicious. A number of medical people have recently questioned the advisability of such scans, due in part to the radiation they impart. Typically, one such scan gives a dose of 12 mSv, applied to the whole body. By contrast, a chest x-ray typically administers 0.20 mSv to only 6.0 kg of tissue. How many chest x rays would deliver the same total amount of energy to the body of a 85 kg person as one whole-body scan?
Physics
1 answer:
aksik [14]3 years ago
7 0

Answer:

Explanation:

Given

One CT scan gives a dose of 12 mSv

and a chest X ray typically administer 0.2 mSv to 6 kg tissue

mass of Person m=85 kg

therefore \frac{85}{6}=14.167

N no of x-Ray equals to 12 mSv dose of CT scan

n=\frac{12}{0.2}=60

Therefore total no of x ray required is =60\times 14.167=850.02\approx 850 X rays

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1. An asphalt block has a mass of 90 kg and a volume of 0.075 m. Determines the density of the asphalt.
Kisachek [45]

Answer:

1. Density = 1200[kg/m^3]; 2. Volume= 0.005775[m^3], mass= 15.59[kg]

Explanation:

1. We know that the density is defined by the following expression.

Density = \frac{mass}{volume} \\where:\\mass=90[kg]\\volume=0.075[m^{3} ]\\density=\frac{90}{0.075} \\density=1200[\frac{kg}{m^{3} }]

2. First we need to convert the units to meters.

wide = 35[cm] = 35/100 = 0.35[m]

long = 11 [dm] =  11 decimeters = 11/10 = 1.1[m]

Thick = 15[mm] = 15/1000 = 0.015[m]

Now we can find the density using the expression for the density.

density= \frac{mass}{volume} \\where:\\volume = wide*long*thick\\volume=0.35*1.1*0.015 = 0.005775[m^3]\\\\mass= density*volume = 2700*0.005775 = 15.59[kg]

7 0
3 years ago
A car is on a circular off ramp of an interstate and is traveling at exactly 25 mph around the curve. Does the car have velocity
netineya [11]

Answer:

The car has velocity and acceleration but is not decelerating

Explanation:

Since the car is traveling at 25 mph around the curve, it has a tangential velocity. This tangential velocity is constantly changing in direction (so the car could adapt to the curve and not moving forward in a straight line), there should be a centripetal acceleration in play here. This acceleration does not slow down the car so it's not decelerating.

6 0
2 years ago
Car A starts out traveling at 35.0 km/h and accelerates at 25.0 km/h2 for 15.0 min. Car B starts out traveling at 45.0 km/h and
lawyer [7]

1 kilometre=1000 metre

      1 hour = 3600 second

       1\ km/hr=\frac{1000}{3600} m/s

       1\ km/hr=\frac{5}{18} m/s

The initial velocity of car A is 35.0 km/hr i.e

                                         35.0\ km/hr=35*\frac{5}{18} m/s

                                                                   = 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as  25\ km/hr^2

                                            =\ 25*\frac{1000}{3600*3600} m/s^2

                                            =0.00192901234 m/s^2

The time taken by car A = 15 min.

From equation of kinematics we know that-

                                 v= u+at      [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A,  v = 9.72 m/s +[0.00192901234×15×60]m/s

                                   =11.456111106 m/s

The acceleration of B is given as    15\ km/hr^2

                                    =0.00115740740740 m/s^2

The time taken by car B =20 min

The final velocity of B is -

                             v= u+at

                               = u-at    [Here a is negative due to deceleration]

                               =12.5 m/s +[0.0011574074074×20×60]

                               =13.8888888.....

                               =13.9

The acceleration of C is given as    40\ km/hr^2          

                                                            =\ 0.003086419753 m/s^2

The time taken by car C =30 min

The final velocity of C is-

                                v = u+at

                                   =8.89 m/s+[0.003086419753×30×60] m/s

                                   =14.4455555555..m/s

                                   =14.45 m/s

The car C is decelerating.The deceleration is given as-  60\ km/hr^2

                                                                      =0.0046296296296m/s^2

The time taken by car D= 45 min.

The final velocity of the car D is-

                     v =u+at

                        =30.56 -[0.00462962962962×45×60]m/s

                        =18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.

                 


3 0
3 years ago
Assignment 10 Coulombic Equation Practice Directions: Complete the following problems to calculate the electrostatic force that
Tatiana [17]

Answer:

Magnitude of the force between the charges is F = 1.92×10^20N

Explanation:

Given the magnitude of force according to coulombs law

F =K[(q1*q2)/r2]

Where q1 and q2 are the charges

r is the distance between the charges

K is the coulombs constant

Substituting the given values, we have;

F = 8.98×10^9 × 1.5×10^6 × 3.2×10^4/1.5²

F = 43.1×10^19/2.25

F = 19.16×10^19N

F = 1.92×10^20N

8 0
3 years ago
Gas a bG1 5.22 0.0289G2 1.05 0.0388G3 2.31 0.0467G4 4.05 0.0310Based on the given van der Waals constants for four hypothetical
inysia [295]

Answer:

Gas 2, Gas 3, Gas 4, Gas 5 is the order of decreasing strength of inter-molecular forces.

Explanation:

The strength increases as there is a decrease in the vanderwaals constant and vice versa.

3 0
3 years ago
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