Answer:
v = 1.076 m /s
Explanation:
Initial volume of balloon = 4/3 x 3.14 x (9.905/2)³
=508.56 m³
Final volume of balloon = 4/3 x 3.14 x (16.502/2)³
= 2351.73 m³
Increase in volume = 1843.17 m³
Cross sectional area of inlet A = 3.14 x( 1.458/2)²
A = 1.6687 m²
Volume rate of flow of air = cross sectional area x velocity of inflow
= 1 .6687 V [ V is velocity of inflow ]
Total time taken = Increase in volume / rate of flow of air
17.108 X 60 = 1843.17 / 1.6687 V
V = ![\frac{1843.17}{1.6687\times17.108\times60}](https://tex.z-dn.net/?f=%5Cfrac%7B1843.17%7D%7B1.6687%5Ctimes17.108%5Ctimes60%7D)
v = 1.076 m /s
Answer:
True. 18-year old teenagers that practice some sport or have a hobby have less references to substance abuse or violence.
Explanation:
When someone practices a certain sport, he/she is occupied with its sport, also, sports encourage responsability, discipline and takes most of the person´s time and energy. People who doesn´t practice sports or have any hobby have more idle time and is always looking for something to take away the boredom. Most of the cases they start doing drugs or start having violent episodes because is what´s more handful and gives them the dopamine that the sport is not giving them (because they don´t practice any).
This is way the display of sports or hobbies involvements is associated with decreased references to risky behavior.
Answer:
ummmmmmmmmm
cause only the smartest can figure it out
Explanation:
i only guessed -.-ll
Answer:
The right answer is "5.105×10⁸ m³/sec".
Explanation:
The given values are:
Catchment area,
A = 3 km²
Length to watershed,
L = 1100 m
Average watershed slope,
S = 0.9% i.e., 0.009
Curve number,
CN = 60
Rainfall duration,
D = 20 min
Let,
- Time form beginning of the rainfall will be "
". - Lag time will be "
".
Now,
⇒ ![t_1=\frac{L^{0.8}\times (\frac{1000}{CN} -9)^{0.7}}{19000S^{0.5}}](https://tex.z-dn.net/?f=t_1%3D%5Cfrac%7BL%5E%7B0.8%7D%5Ctimes%20%28%5Cfrac%7B1000%7D%7BCN%7D%20-9%29%5E%7B0.7%7D%7D%7B19000S%5E%7B0.5%7D%7D)
On substituting the values, we get
⇒ ![t_1=\frac{1100^{0.8}\times (\frac{1000}{60} -9)^{0.7}}{19000\times 0.009^{0.5}}](https://tex.z-dn.net/?f=t_1%3D%5Cfrac%7B1100%5E%7B0.8%7D%5Ctimes%20%28%5Cfrac%7B1000%7D%7B60%7D%20-9%29%5E%7B0.7%7D%7D%7B19000%5Ctimes%200.009%5E%7B0.5%7D%7D)
⇒ ![=0.625 \ hours](https://tex.z-dn.net/?f=%3D0.625%20%5C%20hours)
then,
⇒ ![T_p=\frac{D}{2}+t_1](https://tex.z-dn.net/?f=T_p%3D%5Cfrac%7BD%7D%7B2%7D%2Bt_1)
⇒ ![=\frac{0.33}{2}+0.625](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.33%7D%7B2%7D%2B0.625)
⇒ ![=\frac{0.33+1.25}{2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.33%2B1.25%7D%7B2%7D)
⇒ ![=\frac{1.58}{2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1.58%7D%7B2%7D)
⇒ ![=0.79 \ hr](https://tex.z-dn.net/?f=%3D0.79%20%5C%20hr)
hence,
The peak flow will be:
⇒ ![Q_p=\frac{484A}{t_p}](https://tex.z-dn.net/?f=Q_p%3D%5Cfrac%7B484A%7D%7Bt_p%7D)
⇒ ![=\frac{484\times 3}{0.79}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B484%5Ctimes%203%7D%7B0.79%7D)
⇒ ![=\frac{1452}{0.79}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1452%7D%7B0.79%7D)
⇒ ![=1837.97 \ km^3/hr](https://tex.z-dn.net/?f=%3D1837.97%20%5C%20km%5E3%2Fhr)
or,
⇒
Answer:
Technician A is correct
Explanation:
Spot welding is the type of welding done to join two or more metals together and it is done by applying pressure and heat to the weld area using copper alloy electrodes
Hence Technician A is absolutely correct