Answer:
<em>Linus needs to take one of the zero out and it should be 30 instead 300.</em>
Explanation:
It is because Linus put three zero instead two zero.
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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.
To know more about kVp visit:-
brainly.com/question/17095191
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Answer:
The thrust of the engine calculated using the cold air is 34227.35 N
Explanation:
For the turbofan engine, firstly the overall mass flow rate is considered. The mass flow rate is given as
Here
- ρ is the density which is given as
- P is the pressure of air at 5500 m from the ISA whose value is 50506.80 Pa
- R is the gas constant whose value is 286.9 J/kg.K
- T is the temperature of the inlet which is given as 253 K
- A is the cross-sectional area of the inlet which is given by using the diameter of 2.0 m
- V_a is the velocity of the aircraft which is given as 250 m/s
So the equation becomes
Now in order to find the flow from the fan, the Bypass ratio is used.
Here BPR is given as 8 so the equation becomes
Now the exit velocity is calculated using the total energy balance which is given as below:
Here
- h_4 and h_5 are the enthalpies at point 4 and 5 which could be rewritten as and respectively.
- The value of T_4 is the inlet temperature which is 253 K
- The value of T_5 is the outlet temperature which is 233K
- The value of c_p is constant which is 1005 J/kgK
- V_a is the inlet velocity which is 250 m/s
- V_e is the outlet velocity that is to be calculated.
So the equation becomes
Rearranging the equation gives
Now using the cold air approach, the thrust is given as follows
So the thrust of the engine calculated using the cold air is 34227.35 N