Recent content by destinc

sorry, balloon is rising at constant rate of 1.8m/s. balloon height is 3.1m. for the formula I used (1.8m/s)^2=Vi + 2(-9.8m/s^2)3.1m

A hot-air balloon has just lifted off and is rising at the constant rate of 1.8 . Suddenly one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward. If the passenger is 3.1 above her friend when the camera is tossed, what is the...

1. A dog runs back and forth between its two owners, who are walking toward one another. The dog starts running when the owners are 10.0m apart. If the dog runs with a speed of 3.0m/s, and the owners each walk with a speed of 1.3m/s , how far has the dog traveled when the owners meet? My...

which is the best leaving group? Cl- enolate NR3 I couldn't figure out how to draw the enolate here, but it is a basic 3 carbon one. I believe the answer is Cl because resonance on the enolate will disperse the neg. charge. Am I on the right track or is there a factor I am missing?

so W=70kg(9.81)(.2)(1576)=216447.84 P=216447.84/573= 377.7W Close but not 380W Where is the discrepancy?

A man runs up 1576 steps in 9minutes 33sec. If the height gain of each step was 0.20m, and the mass of the runner was 70.0kgs, what was average power output during climb? I used P=W/T & W=FD W=70kg(1576)0.20=22,064J P= 22,064J/573s = 38.5 W answer given is380W What am I missing?

ahhhhh, Typo is in book. Thanks for the help

I really don't understand the format of the given answer, could one of you translate, why don't they give value of C in the answer?

Jogger A KE= 1/2mv^2 Jogger B KE= 2.25mv^2 Jogger C KE= 3/8mv^2 Jogger D KE=1/2mv^2 Oh C< D=A < B Still not the answer Thanks again for all this help

Jogger C KE=1/2(3m(v/2)^2 = 3/2m(v^2)/4 = 3/8mv^2 Jogger D KE= 1/2(4m)(v/2)^2 = 1/2mv^2 Which means D=A and both > C > B Which still isn't the given answer

[b]Jogger A has a mass m and a speed v, jogger B has a mass m/2 and a speed 3v, jogger C has a mass 3m and a speed v/2 and jogger D has a mass 4m and a speed v/2. Rank the joggers in order of increasing kinetic energy. Indicate ties where appropriate I started with KE formula, KE=1/2mv^2 and...

nvrmind previous post about RHL, I see my mistake. So, for the second part, I get lim (x^2 - 2x +1)= 0 x→1- lim lnx= 0 x→1+ Which means f(x0 is differentiable at 1 and final answer is f(x) is differentiable for all real numbers x≠0 Correct?

Thanks, you are correct about the setup of f(x). I still get 1/4 when I plug 1/2 into x^2 - 2x + 1 1/2^2= 1/4, 2(1/2)=1 1/4 -1 + 1= 1/4 The limits set up for 1 helps, for some reason I thought I couldn't use the polynomial, but I finally found it in my notes. Thank you

f(x)= x=1/x-1 if x ≤ 0 x^2-2x +1 if 0 < x < 1 ln x if x ≥ 1 Here's what I have so far lim f(x)= lim x+1/x-1= -1, LHL= -1 x →0- x→0- lim f(x)= lim x^2-2x+1= (1/2)^2 - [(2)1/2] +1= 1/4, RHL= 1/4 x→0+ x→0+...

thanks, I made the problem so much harder than I needed to. The redirect helped.