Answer:
<h2>698.3Kpa</h2>
Explanation:
Step one:
given data
V1=0.25m^3
T1=290k
P1=100kPa
V2=0.5m^2
T2=405k
P2=? final pressure
Step two:
The combined gas equation is given as
P1V1/T1=P2V2/T2
Substituting we have
(100*0.25)/290=P2*0.05/405
25/290=0.5P2/405
0.086=0.05P2/405
cross multiply
0.086*405=0.05P2
34.9=0.05P2
divide both sides by 0.05
P2=34.9/0.05
P2=698.3Kpa
<u>Therefore the new pressure is 698.3Kpa when the gas is compressed</u>
Answer:
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Explanation:
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Answer:
Option (d) 2 min/veh
Explanation:
Data provided in the question:
Average time required = 60 seconds
Therefore,
The maximum capacity that can be accommodated on the system, μ = 60 veh/hr
Average Arrival rate, λ = 30 vehicles per hour
Now,
The average time spent by the vehicle is given as
⇒ 
thus,
on substituting the respective values, we get
Average time spent by the vehicle = 
or
Average time spent by the vehicle = 
or
Average time spent by the vehicle = 
or
Average time spent by the vehicle =
hr/veh
or
Average time spent by the vehicle =
min/veh
[ 1 hour = 60 minutes]
thus,
Average time spent by the vehicle = 2 min/veh
Hence,
Option (d) 2 min/veh
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
#SPJ4
Answer:
QPSK: 7.5 MHz
64-QAM:2.5 MHz
64-Walsh-Hadamard: 160 MHz
Explanation:
See attached picture.