Answer:
The car's position at 3.0 seconds is 17.5 meters. The car will reach 0.0 meters at 10 seconds.
Explanation:
You can rewrite the equation as P= -2.5t+25. t=time and P is equal to position; so you will plug in 3.0 seconds into t. The equation will now look like: P = -2.5(3.0)+25
P = -2.5(3.0) + 25
P = -7.5 + 25
P = 17.5 meters
Now, to find when the car reaches 0.0 meters, we will plug that into P since P = position and will solve for t.
P = -2.5t + 25
0.0 = -2.5t + 25
-25 -25
-25 = -2.5t
-25/-2.5 = -2.5t/-2.5
10 = t
So the car will reach 0.0 meters at 10 seconds.
Answer:
The rate of irreversible loss will be "55.22 MW".
Explanation:
The given values are:
Elevation,
h = 120 m
Flow of water,
Q = 100 m³/s
Efficiency,
= 80%
i.e,
= 0.8
Efficiency turbine,
= 50 MW
Now,
Without any loss,
The power generated by turbine will be:
⇒ ![P=\delta gQh](https://tex.z-dn.net/?f=P%3D%5Cdelta%20gQh)
On substituting the values, we get
⇒ ![=1000\times 9.8\times 100\times 120](https://tex.z-dn.net/?f=%3D1000%5Ctimes%209.8%5Ctimes%20100%5Ctimes%20120)
⇒ ![=117.72 \ MW](https://tex.z-dn.net/?f=%3D117.72%20%5C%20MW)
Power generated in actual will be:
= ![\frac{50}{0.8}](https://tex.z-dn.net/?f=%5Cfrac%7B50%7D%7B0.8%7D)
= ![62.5 \ MW](https://tex.z-dn.net/?f=62.5%20%5C%20MW)
Hence,
Throughout the piping system,
The rate of irreversible loss is:
= ![Power \ generated \ by \ turbine-Power \ generated \ in \ actual](https://tex.z-dn.net/?f=Power%20%5C%20generated%20%5C%20by%20%5C%20turbine-Power%20%5C%20generated%20%20%5C%20in%20%5C%20actual)
= ![117.72-62.5](https://tex.z-dn.net/?f=117.72-62.5)
= ![55.22 \ MW](https://tex.z-dn.net/?f=55.22%20%5C%20MW)
Answer:
Velocity boundary layer thickness over an internal or internal fixed flat plate is normal to surface distance required for the fluid velocity to attain the 99% of the free stream velocity value.
Explanation:
The affirmation is the current definition of velocity boundary layer thickness.
Answer:
q=2313.04![W/m^2](https://tex.z-dn.net/?f=W%2Fm%5E2)
T=690.86°C
Explanation:
Given that
Thickness t= 20 cm
Thermal conductivity of firebrick= 1.6 W/m.K
Thermal conductivity of structural brick= 0.7 W/m.K
Inner temperature of firebrick=980°C
Outer temperature of structural brick =30°C
We know that thermal resistance
![R=\dfrac{t}{KA}](https://tex.z-dn.net/?f=R%3D%5Cdfrac%7Bt%7D%7BKA%7D)
These are connect in series
![R=\left(\dfrac{t}{KA}\right)_{fire}+\left(\dfrac{t}{KA}\right)_{struc}](https://tex.z-dn.net/?f=R%3D%5Cleft%28%5Cdfrac%7Bt%7D%7BKA%7D%5Cright%29_%7Bfire%7D%2B%5Cleft%28%5Cdfrac%7Bt%7D%7BKA%7D%5Cright%29_%7Bstruc%7D)
Heat transfer
![Q=\dfrac{\Delta T}{R}](https://tex.z-dn.net/?f=Q%3D%5Cdfrac%7B%5CDelta%20T%7D%7BR%7D)
So heat flux
q=2313.04![W/m^2](https://tex.z-dn.net/?f=W%2Fm%5E2)
Lets temperature between interface is T
Now by equating heat in both bricks
![\dfrac{980-T}{\dfrac{0.2}{1.6A}}=\dfrac{T-30}{\dfrac{0.2}{0.7A}}](https://tex.z-dn.net/?f=%5Cdfrac%7B980-T%7D%7B%5Cdfrac%7B0.2%7D%7B1.6A%7D%7D%3D%5Cdfrac%7BT-30%7D%7B%5Cdfrac%7B0.2%7D%7B0.7A%7D%7D)
So T=690.86°C
Answer:
![BW=\Delta f= f_H-f_L=4300Hz-300Hz=4000Hz=4kHz](https://tex.z-dn.net/?f=BW%3D%5CDelta%20f%3D%20f_H-f_L%3D4300Hz-300Hz%3D4000Hz%3D4kHz)
Explanation:
Bandwidth is the difference between the upper and lower frequencies in a continuous set of frequencies. The bandwidth may be determined by use of the following formula:
![BW=f_H-f_L\\\\Where:\\\\f_H=Upper\hspace{3}cutoff\hspace{3}frequency=4.3kHz=4300Hz\\f_L=Lower\hspace{3}cutoff\hspace{3}frequency=300Hz](https://tex.z-dn.net/?f=BW%3Df_H-f_L%5C%5C%5C%5CWhere%3A%5C%5C%5C%5Cf_H%3DUpper%5Chspace%7B3%7Dcutoff%5Chspace%7B3%7Dfrequency%3D4.3kHz%3D4300Hz%5C%5Cf_L%3DLower%5Chspace%7B3%7Dcutoff%5Chspace%7B3%7Dfrequency%3D300Hz)
Hence the signal’s frequency bandwidth is:
![BW=4300-300=4000Hz=4kHz](https://tex.z-dn.net/?f=BW%3D4300-300%3D4000Hz%3D4kHz)