-0 m/s
- average velocity=displacement/time
- the runners displacement is zero so her average velocity must be zero
Answer:
1
Explanation:
A hose has been clamped so that the area at the clamp is only one quarter the area of the rest of the hose. When we ignore the viscosity of water, the ratio of the volume of water delivered per unit time when the clamp is on to the volume of water delivered per unit time without the clamp is 1 as continuity says the same amount of water must flow out
Answer:
"Energy deficiency, no coal-burning, no-cost mining pollution" is the correct answer.
Explanation:
- “The greenest kilowatt-hour seems to be the one this really doesn't should use,” explained Joe Stepenovitch, co-owner as well as COO of something like the electricity IQ Group. Whether a kilowatt becomes generated is far less essential instead of not needing to do something with it.
- It, therefore, reduces operational costs, appeals to progressives and green-conscious consumers, prepares the business for impending emissions reductions policy caps, as well as coincides with you including an imminent future focused on renewable energy sources.
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
1110 atm
Let's start by calculating how many cm deep is 36,000 feet.
36000 ft * 12 in/ft * 2.54 cm/in = 1097280 cm
Now calculate how much a column of water 1 cm square and that tall would mass.
1097280 cm * 1.04 g/cm^3 = 1141171.2 g/cm^2
We now have a number using g/cm^2 as it's unit and we desire a unit of Pascals ( kg/(m*s^2) ).
It's pretty obvious how to convert from g to kg. But going from cm^2 to m is problematical. Additionally, the s^2 value is also a problem since nothing in the value has seconds as an unit. This indicates that a value has been omitted. We need something with a s^2 term and an additional length term. And what pops into mind is gravitational acceleration which is m/s^2. So let's multiply that in after getting that cm^2 term into m^2 and the g term into kg.
1141171.2 g/cm^2 / 1000 g/kg * 100 cm/m * 100 cm/m = 11411712 kg/m^2
11411712 kg/m^2 * 9.8 m/s^2 = 111834777.6 kg/(m*s^2) = 111834777.6 Pascals
Now to convert to atm
111834777.6 Pa / 1.01x10^5 Pa/atm = 1107.2750 atm
Now we gotta add in the 1 atm that the atmosphere actually provides (but if you look closely, you'll realize that it won't affect the final result).
1107.274 atm + 1 atm = 1108.274 atm
And finally, round to 3 significant figures since that's the accuracy of our data, giving 1110 atm.
