Posted by **Karen** on Saturday, April 5, 2008 at 3:24am.

The second hand and the minute hand on one type of clock are the same length L.

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(a) What is the period Tsecond of the motion of the second hand? (_) The period is 10 seconds.

(o) The period is 60 seconds.

(_) The period is 30 seconds.

(_) The period depends on the value for the length L.

(_) The period is 1 second.

Correct. The period is the time it takes for the second hand to make one complete revolution.

(b) What is the period Tminute of the motion of the minute hand? (_) The period is 600 seconds.

(o) The period is 3600 seconds.

(_) The period is 60 seconds.

(_) The period depends on the value for the length L.

(_) The period is 1800 seconds.

Correct. The period is the time it takes for the minute hand to make one complete revolution.

(c) What is the algebraic expression for the ratio ac,second/ac,minute of the centripetal accelerations for the tips of the second hand and the minute hand? Express your answer in terms of the periods Tsecond and Tminute. (Answer using T_m to be the period of the second hand and T_s to be period of the minute hand.)

ac,second/ac,minute =

not sure what algebraic expression means

(d) What is the ratio ac,second/ac,minute of the centripetal accelerations for the tips of the second hand and the minute hand?

ac,second/ac,minute = I'm thinking the ratio would be T_m/T_s. I sthis correct?

I was able to answer the multiple question part of the problem, but not sure what it is asking me to do about the algebraic expression and ratio

- Physics -
**drwls**, Saturday, April 5, 2008 at 9:47am
c) An agebraic expression is a formula. In this case it is

a = V^2/R = (2 pi R/T)^2/R

or

a = 4 pi^2 R/T^2

Where R is the length of the clcok's hand.

(4) The are asking for an algebraic expression for the ratio

(ac,second)/(ac,minute)

Assuming the lengths R of the two hands are the same, that would be

= (T_m/T_s)^2

because the acceleration is inversely proportional to T^2

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