Let point A be 0.0 miles (first city)
Let point B be 160.5 miles (first city to second city)
Let point C be 28.5 miles (first city to mail stop)
Take C – A = C [28.5 - 0.0 = 28.5] (This checks the distance between city 1 and Mail stop)
Then Take B – C = Distance from the first city to the second city [160.5 - 28.5 = 132 Miles]
Answer: The Mail stop is 132 miles from the Second City.
2 times 10 to the power of 5 is your answer.
In a block and tackle, some friction in the pulleys will reduce the mechanical advantage of the machine. To include friction in a calculation of the mechanical advantage of a block and tackle, divide the weight of the object being lifted by the weight necessary to lift it.
Hope this helps
Answer:
0.368 cm
Explanation:
x = distance by which the mercury rise
d = depth of the water = 10 cm = 0.10 m
ρ = density of water = 1000 kgm⁻³
ρ' = density of mercury = 13600 kgm⁻³
P₀ = atmospheric pressure
Using equilibrium of pressure on both side
P₀ + ρ g d = P₀ + ρ' g (2x)
(1000) (0.10) = (13600) (2x)
x = 0.00368 m
x = 0.368 cm
