Answer:
<em>55%</em>
Explanation:
hot reservoir = 1100 K
cold reservoir = 500 K
<em>This is a Carnot system</em>
For a Carnot system, maximum efficicency of the system is given as
Eff = 1 - 
where Tc = temperature of cold reservoir = 500K
Th = temperature of hot reservoir = 1100 K
Eff = 1 - 
Eff = 1 - 0.45 = 0.55 or<em> 55%</em>
Answer: N has to be lesser than or equal to 1666.
Explanation:
Cost of parts N in FPGA = $15N
Cost of parts N in gate array = $3N + $20000
Cost of parts N in standard cell = $1N + $100000
So,
15N < 3N + 20000 lets say this is equation 1
(cost of FPGA lesser than that of gate array)
Also. 15N < 1N + 100000 lets say this is equation 2
(cost of FPGA lesser than that of standardcell)
Now
From equation 1
12N < 20000
N < 1666.67
From equation 2
14N < 100000
N < 7142.85
AT the same time, Both conditions must hold true
So N <= 1666 (Since N has to be an integer)
N has to be lesser than or equal to 1666.
The pressure difference across the sensor housing will be "95 kPa".
According to the question, the values are:
Altitude,
Speed,
Pressure,
The temperature will be:
→ ![T = 15.04-[0.00649(9874)]](https://tex.z-dn.net/?f=T%20%3D%2015.04-%5B0.00649%289874%29%5D)
→ 
→ 
now,
→ ![P_o = 101.29[\frac{(-49.042+273.1)}{288.08} ]^{(5.256)}](https://tex.z-dn.net/?f=P_o%20%3D%20101.29%5B%5Cfrac%7B%28-49.042%2B273.1%29%7D%7B288.08%7D%20%5D%5E%7B%285.256%29%7D)
→ 
hence,
→ The pressure differential will be:
= 
= 
Thus the above solution is correct.
Learn more about pressure difference here:
brainly.com/question/15732832
Answer:
Yes, it is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water that is entering at 20 °C because this temperature (20 °C) of the external cooling water is less than the saturation temperature of steam which is which is 45.81 °C, and heated by a boiler; as a result of this condition, coupled with the assumption that the turbine, pump, and interconnecting tube are adiabatic, and the condenser exchanges its heat with the external cooling river water, it possible to maintain a pressure of 10 kPa.
Answer:
a)Δs = 834 mm
b)V=1122 mm/s
Explanation:
Given that
a)
When t= 2 s
s= 114 mm
At t= 4 s
s= 948 mm
So the displacement between 2 s to 4 s
Δs = 948 - 114 mm
Δs = 834 mm
b)
We know that velocity V
At t= 5 s
V=1122 mm/s
We know that acceleration a
a= 90 t
a = 90 x 5
