Recent content by torresmido

so the plane's acceleration is 11.8 m/s^2 downward. We should sum the gravitation acceleration and the pretzels.

Homework Statement Your airplane is caught in a brief, violent downdraft. To your amazement, the pretzels on your seatback tray rise vertivally, and you estimate their upward acceleration relative to the plane at 2 m/s^2. What's the donward acceleration of the plane? ANy suggestions are...

Thanks guys! this was so helpful

A spring with spring constant k=340 N/m is used to weigh a 6.7 KG fish. How does the spring stretch? I used Hooke's law (F=-Kx), but ended up having a negtive distance x=-0.2m. Is this expected? in the problem they say the spring stretches...this is confusing me...thank you for your help guys...

For player 1: x=0.25t^2 (this player is just accelerating) when you subtitute t by 7.5 you ger 14...as expected... Thank you for the questions...it turns out to be a good preparation for my first physics test

I can do that, but I'm confused about one thing. When they that the vector is pointing to the right, does it mean that its direction is 0 degrees from the positive x-axis or its direction is some angle between the the x-axis and the y-axis?

Homework Statement Vector A points to the right with magnitude 10 m/s, and vector B points upward vertically with magnitute of 15 m/s. What is the magnitude and direction of vector c such that A+B+C=0 Any ideas on this problem are appreciated.. Homework Equations The Attempt...

For the second vector the vector in the ight of the picture), what angle did you use? The angle should be 180°-22° = 158°

Sorry guys ... I see my mistake now. the limimit is 2/e so the series diverges by the ratio test

Alright, the series you have is n!*2^n*n^-n this can be simplifyed to become: An= n!*(2/n)^n so An+1= (n+1)!*(2/n)^(n+1) "you just replace n by n+1 The ratio test says lim as n goes to infinity of An+1/An which is the same as: lim as n aprroaches infiniti of...

This is a free-fall problem: the equation to be used is, x=0.5*a*t^2 since a=g=9.8 m/s x=0.5*9.8*t^2 = 4.9*t^2 you have t=6.5s so x=4.9*9.8*(6.5)^2 = 207 meters. which is how deep the mineshaft. Free free to ask any other questions

dS/dt=kS*[1-(S/N)]*[(S/M)-1] Assume that K ans M are constans (where M is lower or Equal to N). Find the bifurcation value for N? I really didn't know where to start. Any help is appreciated

(2n+2)/n =n*(2+2n^-1)/n so now you cancel out the n in the denominator with the one above it you get: 2+(2/n) when you take the limit as n approaches infiniti you get 2

Thanks a lot...YOU SAVED MY LIFE...I got everything you said ;)

you should first make the series simple: The series you have is the same as: the sum from n=1 to infiniti of n!*(2/n)^n "because n^(-n) is the same as 1/n^(n)). Now you can use the ratio test: take the limite as n approaches infiniti of [(n+1)!*(2/n)^(n+1)] / [n!*(2/n)^n ] this limit...