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Camera A has a lens with an aperture diameter of 8.00mm . It photographs an object using the correct exposure time of 3.33×10−2s .

What exposure time should be used with camera B in photographing the same object with the same film if this camera has a lens with an aperture diameter of 22.5mm ? The lenses of cameras A and B have the same focal length.

I'd like to say I attempted this, but I don't even know which equation to use. My book doesn't have an equation relating exposure time to focal length or diameter.

(My physics final is tomorrow)

Okay I need some help with this physics problem:

A pendulum clock has a heavy bob supported on a very thin steel rod that is 1.00000 m long at 22.0 deg C.

Part A) Find the period (T) to 6 sig figs. I correctly answered T = 2.00709

Part B) To 6 significant figures, what is the clock's period if the temperature increases by 13.0 deg C? Assume that coefficient of linear expansion of steel is 1.20×10^-5/K I correctly answered T = 2.0725s

Part C) The clock keeps perfect time at 22.0 deg C. At 35.0 deg C after how many hours will the clock be off by 3.00s ?

I subtracted 2.00725-2.00709 = .00016sec

3sec/.00016sec = 18750sec \ 3600sec = 5.2083 hours. My answer was wrong.

Any insight on how to do this?

Long story short, I'm working on a short story that takes place in Russia in the 1950's (ish). I know that the Russian currency is the rouble and kopeks are less than roubles, but I'm looking for a Russian phrase that is used to call someone cheap (if there is one). I realize penny pincher doesn't work, but I didn't want to simply switch to "kopek pincher" seeing as it sounds ridiculous.

Obviously, I'd like the translated Russian phrase.

Evaluate the integral 1/(1-x^2) dx to 3 decimal places from 0 to 1/8 [0,1/8] are the boundaries.

The boundaries make it a proper integral, so I don't think I'm suppose to take the limit. I was tempted to substitute arctan(x) in there, but I'm not sure if that will work, or how. The problem didn't specify whether or not to use approximation, so I'm assuming there's another way? :-( Help please.