Recent content by PhysicsAdvice

Need to find the angle at which a plane should aim if when traveling at 210km/h with a 40km wind east produces a resultant angle of 60 degrees. I can write that 60=tan^-1(210sin60/(210cos60+40) and proceed from there: tan60=210sinx/(210cosx+40) 1.73=210sinx/210cosx+40...

found the right one, thanks!

In a locket around my neck, of course, let me check

How would i solve sinxcosx=0.458, I cannot determine how to isolate the angle, is there a set of identities i could use? Thanks!

i got it! thanks a lot i just didn't see the equivalency thanks!

f(x)= -3x^4/(4x-8)^(1/2), f'(x)=? Cannot figure out how to do this one, would appreciate help finding my mistakes!I'm sure you all cna find the answer yourself but for convenience it is -3x^3(7x-16)/(4(x-2)^(3/2))

having not learned derivatives yet i suppose the method i indicated is the most suitable then?

One of the homework questions i did tonight was (x^5-32)/(x-2), where i had to find the limit when x approaches two. In order to do this i looked up the formula to use for a difference of fifths and was able to solve the question by factoring out the denominator and putting two in for x, giving...

simply, what values of x will produce a positive average profit? I have gotten three questions which I believe to be right wrong in a row now, I believe the answers are off :)

Revenue Equation: R(x)=-x^2+10x Cost Equation: C(x)= 4X+5 Average profit= profit equation, P(x)/x therefore p(x)= R(x)-C(x)=-x^2+6x-5 (-x^2+6x-5)/x=(-1(x-5)(x-1))/x, I then found that x is positive between 1 and 5, therefore average profit is positive in that range, however, the answer...

should the time it takes also be equal since t=v2-v1/a and a=μmg/m?

A truck with a mass of 4500kg and a car with a mass 1500kg both apply the brakes at the same time, if the coefficient of friction is 0.15, and both vehicles are traveling at 20m/s, how far will it take each vehicle to stop? Ff=0.15Fn W=FappD Ek2-Ek1=W Fn=mg Ek=1/2mv^2 for the truck...

right, so at first it would be in one direction then a different direction but does the value change other than that?

if a ball is thrown and caught at the same height with constant horizontal velocity and no vertical displacement is its initial and final velocity the same? since v2y^2=v1y^2+2aDy, no displacement means final vertical velocity will be the same as initial, since v2x=v1x, horizontal velocity...

dumb question, would the normal force be different if the object was pushed rather than pulled?