## Trending News

# What is the value of k and c in this equation? T= k√n +c?

The lab we are doing is a pendulum lab with paper clips as the swinging item. T is the period and n is the number of paper clips.

- I know it's a linear relationship but how would one solve for k with a theoretical set of numbers

- btw i'm trying to extrapolate a theoretical slope and y-intercept but if n=0 would k be undefined?

****also could you please rearrange the equation to solve for k and c respectively. I just want to be sure i have it right.****

what would be the units for k and c if T is in seconds?

### 2 Answers

- gp4rtsLv 71 decade agoFavorite Answer
If you are doing the experiment by adding paper clips and measuring T to get the relation, then n is the independent variable, and the relationship is non-linear. The c is the value of T = T0 when n = 0 (c is not undefined at n = 0). Then for another value of n, say n1, get another value of T = T1 = k√n1 + c.

c = T0

T1 = k√n1 + T0

k = (T1 - T0)/√n1

The units of both c, T and k are seconds, since n is dimensionless.

- pilgrimLv 44 years ago
The coefficients are the hard area. in case you look on the easy circumstances, you get pascal's trianglefor (a + b) ^ok a million (ok=0) a million + a million (ok = a million) a million + 2 + a million (ok = 2) a million + 3 + 3 + a million (ok = 3) a million + 4 + 6 + 4 + a million (ok = 4) a million + 5 + 10 + 10 + 5 + a million ( ok = 5) the ingredient to be conscious is that the coefficients are mixtures of the style combin(ok, n) the place ok is the flexibility, and n is the coefficient. to maintain the typing down, i'm going to apply (ok,n) to signify the mixture of n products from ok products So, for ok = 4: (4,0) + (4, a million) + (4,2) + (4,3) + (4,4) it rather is a million, 4, 6, 4, a million so (a + b) ^4 = a^4 + 4(a^3b) + 6(a^2b^2) + 4(ab^3) + b^4 with a bit of luck, you could take this to the subsequent step. (additionally look up "binomial advance" - it rather is the right call for one among those action)