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oksian1 [2.3K]
3 years ago
13

A 75 kV x-ray machine gives an exposure of 200 mR for a 0.2-sec chest x-ray exam.

Physics
1 answer:
pashok25 [27]3 years ago
7 0

Explanation:

Given that,

Potential = 75 kV

Exposure = 200 mR

Time = 0.2 sec

We need to calculate the x-ray fluence during this chest x-ray exam

Using formula of fluence

x-ray\ fluence=exposure\times time

Put the value into the formula

x-ray\ fluence=200\times0.2

x-ray\ fluence=40\ mRs

We need to calculate the energy fluence

Using formula of energy fluence

E_{f}=40\times10^{-3}\times75\times10^{3}

E_{f} = 3000\ J

We need to calculate dose -equivalent delivered to the bone, muscle, and fat

Using formula of dose

D=\dfrac{E}{t}

Where, D = dose

E = energy

t = time

Put the value into the formula

D=\dfrac{3000}{0.2}

D=15.0\ kJ

Hence, This is the required solution.

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This is the simplified definition.

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Why do you think it is a driving force of weather around the world?
Finger [1]

Answer:

The sun.

Explanation:

The sun provides energy for living organisms, and it drives our planet’s weather and climate patterns.

Remember, Earth is spherical and the energy from the sun does not reach all areas with equal intensity. Areas exposed to the sun are directly on the sun’s rays (i.e. those nearest to the equator) and hence, receive greater solar input. In contrast, those in higher latitudes receive sunlight that is spread over a larger area and that has taken a longer path through the atmosphere. As a result, these higher latitudes receive less solar energy.

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2 years ago
A 0.300 kg ball, moving with a speed of 2.5 m/s, has a head-on collision with at 0.600 kg ball initially at rest. Assuming a per
FrozenT [24]

Answer:

1.25 m/s

Explanation:

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Mass of first ball=0.3 kg

Its speed before collision=2.5 m/s

Its speed after collision=2 m/s

Mass of second ball=0.6 kg

Momentum of 1st ball=mass of the ball*velocity

=0.3kg*2.5m/s

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Momentum of 2nd ball=mass of the ball*velocity

=0.6 kg*velocity of 2nd ball

Since the first ball undergoes head on collision with the second ball,

momentum of first ball=momentum of second ball

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4 0
3 years ago
An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com
solong [7]

Answer:

(a) 3.81\times 10^5\ Pa

(b) 4.19\times 1065\ Pa

Explanation:

<u>Given:</u>

  • T_1 = The first temperature of air inside the tire = 10^\circ C =(273+10)\ K =283\ K
  • T_2 = The second temperature of air inside the tire = 46^\circ C =(273+46)\ K= 319\ K
  • T_3 = The third temperature of air inside the tire = 85^\circ C =(273+85)\ K=358 \ K
  • V_1 = The first volume of air inside the tire
  • V_2 = The second volume of air inside the tire = 30\% V_1 = 0.3V_1
  • V_3 = The third volume of air inside the tire = 2\%V_2+V_2= 102\%V_2=1.02V_2
  • P_1 = The first pressure of air inside the tire = 1.01325\times 10^5\ Pa

<u>Assume:</u>

  • P_2 = The second pressure of air inside the tire
  • P_3 = The third pressure of air inside the tire
  • n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)

Part (a):

Using the above equation for this part of compression in the air, we have

\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa

Hence, the pressure in the tire after the compression is 3.81\times 10^5\ Pa.

Part (b):

Again using the equation for this part for the air, we have

\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa

Hence, the pressure in the tire after the car i driven at high speed is 4.19\times 10^5\ Pa.

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2 years ago
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