can someone please help me with these algebra promlems!..?

Question

Problem 1: The electric resistance R of a wire varies directly as its length L and inversely as the square of its diameter. A wire 20 meters long and 0.6 centimeters in diameter made from a certain alloy has a resistance of 36 ohms. What is the resistance of a piece of wire 60 meters long and 1.2 centimeters in diameter made from the same material? & Problem 2: The period of a simple pendulum is directly proportional to the square root of its length. If a pendulum has a length of 6 feet and a period of 2 seconds, to what length should it be shortened to achieve a 1 second period?

Answer 1

PROBLEM ONE
The electric resistance R of a wire varies directly as its length L and inversely as the square of its diameter. So you mean R=cte*L/d2 where the constant depends on the metal used.
for that alloy, C=Rd2/L=36*0.6*0.6/20=0.648
In the second wire, you use the same constant
R=0.648*60/(1.2*1.2)=27 ohms

PROBLEM TWO
here, period=T=cte*sqrt(L)
cte=T/sqrt(L)=2/sqrt 6=0.816
L=(T/cte)2=(1/0.816)2=1.22*1.22=1.5

Answer 2

Use your windows calculator for #2.

Put in 6 and get the square root...2.45 (rounded up)

Therefore you would have to figure out the square root of half of that.... which is 1.225

Multiply it by itself:

1.225 X 1.225 = 1.500625 (rounded it would be 1.501)

So this is TOTALLY a guess, but for #2 if its multiple choice, I would look for something close to 1.5 feet

Answer 3

(?40)(?fifty 3) (?40) is also written as (?2x2x2x5) taking all the finest factors of 40 besides the undeniable fact that fifty 3 is already a best decision in view that its factors are purely a million and itself, hence (?2x2x2x5)?fifty 3) (?2x2) = 2 hence the answer is two(?10)(?fifty 3) or 2?530)