The x-ray beam's penetrating power is regulated by kVp (beam quality). Every time an exposure is conducted, the x-rays need to be powerful (enough) to sufficiently penetrate through the target area.
<h3>How does kVp impact the exposure to digital receptors?</h3>
The radiation's penetration power and exposure to the image receptor both increase as the kVp value is raised.
<h3>Exposure to the image receptor is enhanced with an increase in kVp, right?</h3>
Due to an increase in photon quantity and penetrability, exposure at the image receptor rises by a factor of five of the change in kVp, doubling the intensity at the detector with a 15% change in kVp.
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Answer:
14.52 minutes
<u>OR</u>
14 minutes and 31 seconds
Explanation:
Let's first start by mentioning the specific heat of air at constant volume. We consider constant volume and NOT constant pressure because the volume of the room remains constant while pressure may vary.
Specific heat at constant volume at 27°C = 0.718 kJ/kg*K
Initial temperature of room (in kelvin) = 283.15 K
Final temperature (required) of room = 293.15 K
Mass of air in room= volume * density= (4 * 5 * 7) * (1.204 kg/m3) = 168.56kg
Heat required at constant volume: 0.718 * (change in temp) * (mass of air)
Heat required = 0.718 * (293.15 - 283.15) * (168.56) = 1,210.26 kJ
Time taken for temperature rise: heat required / (rate of heat change)
Where rate of heat change = 10000 - 5000 = 5000 kJ/hr
Time taken = 1210.26 / 5000 = 0.24205 hours
Converted to minutes = 0.24205 * 60 = 14.52 minutes
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
The attached file gave a detailed solution of the problem.
Answer:
Explanation:
cross sectional area A = 1.9 x 2.6 x 10⁻⁶ m²
= 4.94 x 10⁻⁶ m²
stress = 42 x 9.8 / 4.94 x 10⁻⁶
= 83.32 x 10⁶ N/m²
strain = .002902 / 2.7
= 1.075 x 10⁻³
Young's modulus = stress / strain
= 83.32 x 10⁶ / 1.075 x 10⁻³
= 77.5 x 10⁹ N/m²
Answer:
Temperature
Explanation:
In an ideal gas the specific enthalpy is exclusively a function of Temperature only this can be also written as h = h(T)
A gas is said be ideal gas if obeys PV= nRT law
And in a ideal gas both internal energy and specific enthalpy are a function of Temperature only. Therefore the constant volume and constant pressure specific heats Cv and Cp are also function of temperature only.
