A clock is constructed so that it keeps perfect time when its simple pendulum has a period of 1.000 s at locations where g 5 9.800 m/s2. The pendulum bob has length L 5 0.248 2 m, and instead of keeping perfect time, the clock runs slow by 1.500 minutes per day. (a) What is the free-fall acceleration at the clock’s location? (b) What length of pendulum bob is required for the clock to keep perfect time?

Answer

Given,

Period of the Pendulum, T= 1 s

acceleration due to gravity, g = 9.8 m/s²

Length of Bob = 0.2482 m

Loss of time in the clock = 1.5 minutes/ day

Loss of time in 1 oscillation

   =

   = 0.001042

Hence, time period, T = 1 + 0.001042

                                  T = 1.001042 s

a) Using Time period of oscillation formula

Free fall acceleration at the clock location = 9.77 m/s²

b) Length of pendulum bob to keep it perfect time

Hence, Length of pendulum bob is equal to 0.2475 m.

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3