Objects can have the same mass (but different <span>compositions). Only mass or volume cannot tell you if the object is solid or vo</span>lumes) or same volume (but different masses)
1 volt = 1 joule per coulomb.
Current doesn't actually pass 'through' a battery.
But if it did, then each coulomb would gain or lose 6 joules in traversing 6 volts, depending on its sign, and whether it climbed or fell.
Answer:
(a). 14.4 lbf/in^2.
(b). 27.8 in, AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.
Explanation:
So, from the question above we are given the following parameters which are going to help us in solving this particular Question;
=> The "barometer accidentally contains 6.5 inches of water on top of the mercury column (so there is also water vapor instead of a vacuum at the top of the barometer)"
=> "On a day when the temperature is 70oF, the mercury column height is 28.35 inches (corrected for thermal expansion)."
With these knowledge, let us delve right into the solution;
(a). The barometric pressure = water vapor pressure + acceleration due to gravity (ft/s^2) × water density(slug/ft^3) × {ft/12 in}^3 × [ height of mercury column + specific gravity of mercury × height of water column].
The barometric pressure= 0.363 + {(62.146) ÷ (12^3) × 390.6425}. = 14.4 lbf/in^2.
(b). { (13.55 × length of mercury) + 6.5 } × (62.15÷ 12^3) = 14.4 - 0.603.
Length of mercury = 27.8 in.
AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.
Answer:
If T^2 = a^3
T1^2 / T2^2 = a1^3 / a2^3
Or T2^2 = T1^2 * ( a2^3 / a1^3)
Given T1 = 1 yr and a2 / a1 = 17
Then T2^2 = 1 * 17^3
or T2 = 70 yrs
