Answer:
The maximum acceleration experienced by the sprinter is
and it is experienced in the first phase when the sprinter starts his sprint.
Explanation:
Since the sprinter attains a final speed of 34 km/h in runing 200 meters while starting from rest we have
Using the third equation of kinematics we have
![v^2=u^2+2as](https://tex.z-dn.net/?f=v%5E2%3Du%5E2%2B2as)
where
v is the final speed
u is the initial speed
a is the acceleration
s is the distance covered
Since it is given that sprinter starts from rest thus u= 0 m/s and v= 34 km/h = 9.4m/s this speed is attained at s = 78 m
Applying values in the above equation we get
![a=\frac{v^2-u^2}{2s}=\frac{9.4^2-0^2}{2\times 78}=0.566m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv%5E2-u%5E2%7D%7B2s%7D%3D%5Cfrac%7B9.4%5E2-0%5E2%7D%7B2%5Ctimes%2078%7D%3D0.566m%2Fs%5E2)
Since after that the sprinter moves at a constant velocity thus in that phase it's acceleration is ![0m/s^2](https://tex.z-dn.net/?f=0m%2Fs%5E2)
Now since the sprinter decelerates to 30 km/h or 8.33 m/s in final 63 meters thus the deceleration experienced is again found by third equation of kinematics as
![r=\frac{8.3^2-9.4^2}{2\times 63}=-0.154m/s^2](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B8.3%5E2-9.4%5E2%7D%7B2%5Ctimes%2063%7D%3D-0.154m%2Fs%5E2)
Upon comparing the maximum acceleration experienced by the sprinter is ![0.566m/s^2](https://tex.z-dn.net/?f=0.566m%2Fs%5E2)
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:
Following are the solution to this question:
Explanation:
Whenever a chemical reaction occurs between water and cement the heat is released, and a
(C-S-H gel) gel constructs gel is also recognized as "tobermorite gel."
This one gel acts like a pack of gum and also has a cement quality, that holds its particles intact and therefore contributes to the overall compression mix. An increase in supply explicitly causes the movement in the outcome of power. C3S and C2S are both the compounds of Bouge that produce hydration C-S-H gel.
It mixture must be balanced as
with C-S-H gel also is given as a byproduct. It
, that cause sudden with sulphate and form
, is an unacceptable substance. Sulfate attack or later deterioration of its cement is caused by this
.
All C3S and C2S generate various amounts of C-S-H gel so, the required strength can be maintained without compromising on real term durability.
Answer:
150kW
Explanation:
Energy date of air when it enters= 100kW
Energy rate of air when it leaves= 250kW
Energy rate gained by air= energy rate lost by sytem= 250-100
= 150kW