I believe it's B. the transmission of heat across matter
The type of weather modification is intentional weather modification. Intentional in such a way that there are practical methods applied in order to force/alter the weather condition. For example, frost prevention on crops. Farmers normally utilize large fans to get the warmer air from above to mix with the cold air near the ground surface. By doing so, crops are kept from being ruined by frost.
KNOWN: Long, 30mm-diameter cylinder with embedded electrical heater; power required
to maintain a specified surface temperature for water and air flows.
FIND: Convection coefficients for the water and air flow convection processes, hw and ha,
respectively.
ASSUMPTIONS: Flow is cross-wise over cylinder which is very long in the direction
normal to flow.
The convection heat rate from the cylinder per unit length of the cylinder has
the form
q' = h*(pi*D)*(Ts-Tinf)
and solving for the heat transfer convection coefficient, find
Water
hw = q'/((pi*D)*(Ts-Tinf))
hw = (38*10^3 W/m) / ((pi*(0.030m))*(80-25)C)=
7330.77314 W/m^2K
Air
ha = (400W/m) / ((pi*(0.030m))*(80-25)C)=<span>
77.166033 </span> W/m^2K
COMMENTS: Note that the air velocity is 10 times that of the water flow, yet
hw ≈ 95 × ha.
These values for the convection coefficient are typical for forced convection heat transfer with
liquids and gases
Watter is a better convective heat transfer media than air
Answer:
0.243
Explanation:
<u>Step 1: </u> Identify the given parameters
Force (f) = 5kN, length of pitch (L) = 5mm, diameter (d) = 25mm,
collar coefficient of friction = 0.06 and thread coefficient of friction = 0.09
Frictional diameter =45mm
<u>Step 2:</u> calculate the torque required to raise the load
![T_{R} = \frac{5(25)}{2} [\frac{5+\pi(0.09)(25)}{\pi(25)-0.09(5)}]+\frac{5(0.06)(45)}{2}](https://tex.z-dn.net/?f=T_%7BR%7D%20%3D%20%5Cfrac%7B5%2825%29%7D%7B2%7D%20%5B%5Cfrac%7B5%2B%5Cpi%280.09%29%2825%29%7D%7B%5Cpi%2825%29-0.09%285%29%7D%5D%2B%5Cfrac%7B5%280.06%29%2845%29%7D%7B2%7D)
= (9.66 + 6.75)N.m
= 16.41 N.m
<u>Step 3:</u> calculate the torque required to lower the load
![T_{L} = \frac{5(25)}{2} [\frac{\pi(0.09)(25) -5}{\pi(25)+0.09(5)}]+\frac{5(0.06)(45)}{2}](https://tex.z-dn.net/?f=T_%7BL%7D%20%3D%20%5Cfrac%7B5%2825%29%7D%7B2%7D%20%5B%5Cfrac%7B%5Cpi%280.09%29%2825%29%20-5%7D%7B%5Cpi%2825%29%2B0.09%285%29%7D%5D%2B%5Cfrac%7B5%280.06%29%2845%29%7D%7B2%7D)
= (1.64 + 6.75)N.m
= 8.39 N.m
Since the torque required to lower the thread is positive, the thread is self-locking.
The overall efficiency = 
= 
= 0.243
Answer:
15m
Explanation:
Hello! first to solve this problem we must find that so much that the tree grew in the 15 years this is achieved by dividing the height of the tree before and after

the tree grew 10 times its initial length in 15 years, so to find how tall the nail is, we multiply this factor by 1.5m
X=(1.5m)(10)=15m
the nail is 15 meters above ground level