Recent content by emc92

A brass plug is to be placed in a ring made of iron. At 10°C, the diameter of the plug is 8.731 cm and that of the inside of the ring is 8.719 cm. a) They must both be brought to what common temperature in order to fit? b) What if the plug were iron and the ring brass? formula for...

ohhhhh wow now i see. thanks so much!

(1) Show that |ln 2 - ƩNn=1 ((-1)n-1)(1/n)| ≤ 1/(N+1) (2) Show that |sin x - ƩNn=0 ((-1)n)/(2n+1)!| ≤ |x|2N+2/(2N+2)! I really don't know where to start. should I change the sums to series first then work my way through? Please help!

i just looked over my notes again. we started this problem in class, and our professor told us to pick a value for k above the value of e. so if k is 3, 3/(n+1)! = 0.000005 so now i have to find n?

how do i figure out c?

Use the error estimate for Taylor polynomials to find an n such that | e - (1 + (1/1!) + (1/2!) + (1/3!) + ... + (1/n!) | < 0.000005 all i have right now is the individual components... f(x) = ex Tn (x) = 1/ (n-1)! k/(n-1)! |x-a|n+1 = 0.000005 a = 0 x = 1 I don't know where to go from here

When 22.63 mL of aqueous NaOH was added to 1.226 g of cyclohexylaminoethanesulfonic acid (FM 207.29, structure in the table) dissolved in 41.37 mL of water, the pH was 9.24. Calculate the molarity of the NaOH. How do I start this calculation?

i figured this one out. thanks!

The measured pH of 0.100 M HCl at 25°C is 1.092. From this information, calculate the activity coefficient of H+. I tried using the Debye-Huckel equation and got 0.825 but it's not right. I don't know how else to solve for the activity coefficient.

so for part a, do i need to add 1/2kx^2 to the mgh side to account for the spring being stretched?

i know that if the mass is NOT attached to the spring, the formula to find d (distance up the slope) is 1/2kx^2=mgh. i don't understand how attaching the two together will change it.. and to answer part b, i need the equation to answer part a, right?

A spring (75 N/m) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 2.2 kg is placed at its free end on a frictionless slope which makes an angle of 41 degrees with respect to the horizontal. The spring is then released. a) If the mass is...

i just need some help on how to start the process of proving it. suggestions/recommendations/anything will help!

but if k = 1 - (b^2/a^2) and under the radical says 1-ksin^2(θ), shouldn't the end result under the radical, when expanded, be 1 - (sin(θ))^2- (b^2/a^2)(sin(θ))^2? and where does the 4 outside the integral come from?

wow, i did not see that. thanks so much!