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...
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...
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
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...