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
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
(a) The period Tsecond of the motion of the second hand is 1 second.
(b) The period Tminute of the motion of the minute hand is 3600 seconds.
(c) 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 would be Tsecond/Tminute.
(d) The ratio ac,second/ac,minute of the centripetal accelerations for the tips of the second hand and the minute hand is Tsecond/Tminute.
For part (c), the problem is asking for an algebraic expression for the ratio of the centripetal accelerations for the tips of the second hand and the minute hand. To find this ratio, we need to consider the relationship between the centripetal acceleration and the period of motion.
The centripetal acceleration, ac, is given by the equation ac = (4π^2r)/T^2, where r is the radius (length) of the hand and T is the period of motion.
Let's denote the centripetal acceleration of the second hand as ac_second and the centripetal acceleration of the minute hand as ac_minute.
Since both hands have the same length L, their radii are also the same. Therefore, the ratio of their centripetal accelerations can be expressed as:
ac_second/ac_minute = [(4π^2L)/T_second^2] / [(4π^2L)/T_minute^2]
Simplifying this expression, we get:
ac_second/ac_minute = T_minute^2 / T_second^2
So the algebraic expression for the ratio of the centripetal accelerations is:
ac_second/ac_minute = (T_minute/T_second)^2
For part (d), as you correctly mentioned, the ratio of the centripetal accelerations is given by T_minute/T_second.
For part (c), the question is asking you to find an algebraic expression for the ratio of the centripetal accelerations for the tips of the second hand and the minute hand in terms of their periods.
To do this, you can start by considering the relationship between centripetal acceleration and period. The centripetal acceleration, ac, is given by the formula ac = (4π²r)/T², where r is the radius of the circular motion and T is the period of the motion.
For the second hand, the radius is the length L, and the period is Tsecond. So the centripetal acceleration of the second hand is ac,second = (4π²L)/Tsecond².
For the minute hand, the radius is also L, and the period is Tminute. So the centripetal acceleration of the minute hand is ac,minute = (4π²L)/Tminute².
Now, you need to find the ratio of these two accelerations: ac,second/ac,minute. This can be done by dividing the expression for ac,second by the expression for ac,minute:
ac,second/ac,minute = [(4π²L)/Tsecond²] / [(4π²L)/Tminute²]
Notice that the lengths L cancel out, and you are left with:
ac,second/ac,minute = (Tminute²) / (Tsecond²)
This is the algebraic expression for the ratio of the centripetal accelerations for the tips of the second hand and the minute hand.
For part (d), you are asked to calculate the numerical value of the ratio ac,second/ac,minute. Since you already have the periods Tsecond and Tminute, you can substitute those values into the expression:
ac,second/ac,minute = (Tminute²) / (Tsecond²)
Remember to square the values of the periods correctly before dividing them to find the ratio.