Answer:
In the absence of air resistance. I think no. D ) The bowling ball.
<em><u>Hope</u></em><em><u> this</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
Answer each friend will get 3.33333 repeating if he is included. if only his friends are getting them then each one gets 4
Explanation:
devide 20/6 and 20/5 respectively.
<span>Answer:
The temperature doesn't affect the evaporation rate, but affects on how much of water a parcel of air can contain when saturated which is known by the absolute humidity. Hurricanes are usually happening when the temperature of the sea water west of the Cape Verde islands is over 27 degrees Celsius. If ahead of the path of a hurricane, the sea water temperature drops then it will be less moisture in the air and perhaps the hurricane will fade out. But it is not as simple. How strong a tropical storm is is relative to the difference of temperture between ground level and the top of the troposphere. The greater the difference, the faster the air will rise and the deeper the pressure will be, forcing surrounding air to rush in, thus forming a hurricane force wind. Then there is the fact that the wet adiabatic lapse rate is about half that of dry air. It means that rising moist air cools down slower and therefore rises higher. Hence water is the true fuel of bad weather. But it can't be isolated from the fact that the difference of temperature must be great too. What we often forget is that the tropopause (the border to the stratosphere) is much higher over the equator and therefore, much colder than e.g. the poles.</span>
To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,



So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
Expansion work against constant external pressure: w=-pex Δ Δ V 3. The attempt at a solution . I tried following that. Because Vf>>Vi, and Vf=nRT/pex, then w=-pex x nRT/pex=-nRT (im assuming n is number of moles of CO2?). 1 mole of CaCO3 makes 1 mole of CO2, so plugging in numbers, I get 8.9kJ, although I dont use the 1 atm pressure at all