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:
While In an <em><u>ideal/isolated</u></em> system, as long as the object is not in motion, its potential energy will be the same.
However, <u>potential energy is relative</u>. On Earth, usually, it is measured with respect to gravity. <u>The higher the object, the greater the potential gravitational energy</u>. It's all relative. For the sake of this question, I would assume that potential energy increases.
Explanation:
While kinetic energy depends upon speed, potential energy is always relative to some arbitrary reference point.
Source https://www.physicsforums.com/threads/potential-energy-kinetic-energy.11481/
Weight of the barbell W = 200 Ndistance of the joint is r = 40 cm = 0.4 mtorque created by the weight at the joint is τ = F*r = 200 N*0.4 m = 80 N.mat equilibrium condition , Στ = force*distance - 80 N.m = 0 F'*0.4 - 80 N.m = 0 F'*0.4 = 80 force F' = 200 N
