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.
Answer:
L = 8694 Kg.m²/s
Explanation:
r = 270 ĵ m
v = 14 î m/s
m = 2.3 kg
θ = 90º
L = ?
We can apply the equation
L = m*v*r*Sin θ
L = (2.3 kg)*(14 m/s)*(270 m)*Sin 90º = 8694 Kg.m²/s
I think it’s C but I’m not sure:/
Answer:
500 hours
Explanation:
The sum of total hours over its lifetime will be given by
Where T is total time, r is rate in decimal and To is the original charge hours. Substituting the original charge hours with 5 hours and rate as 0.99 then the time will be
Therefore, the time is equivalent to 500 hours
Answer:
Drag force=1/2Cd*Area*density*V^2
A=240 ft^2
cd=0.75
Drage force=675 lbs
