Recent content by lollikey

my attempt at the solution was delta(x) = 10-2/8 = 1 1/3[ 1+2(1.5)+4(2)+2(2)+4(3)+2(3.5)+4(4)+2(3.5)+1] = 59/3

Homework Statement (a) If the region shown in the figure is rotated about the y-axis to form a solid, use Simpson's Rule with n = 8 to estimate the volume of the solid. (Round your answer to the nearest integer.) Homework Equations delta(x) = b-a/n delta(x)/3 [ f(x) + 4f(x)+ 2f(x) +...

.08k *.26 = .5 * k * x^2 .0208 = .5 * x^2 .0146 = x^2 x = .2039 am I using the wrong height? .08k * h = .5 * k * .26^2 .08 * h = .0338 h = .4225 none of these are right

.08k = .5 * k * x^2 .08= .5 x^2 .16= x^2 x= .40 but it marked it wrong

Homework Statement A spring-like trampoline dips down 0.08 m when a particular person stands on it. If this person jumps up to a height of 0.26 m above the top of the trampoline, how far with the trampoline compress when the person lands? I am very lost! Please help! Homework Equations F=kx...

that is the hint my teacher gave me because of magma

I have been trying to find out of Boyle's Law and plate tectonics are in any way shape or form related. Are they? If they are not, is there a way I can relate their movement to physics?

thank you sooo much! you are the best ever!

9.8msin(theta) - (9.8m)cos(theta) * .7 = 0 9.8msin(theta) = (9.8m)cos(theta) * .7 9.8msin(theta) /(9.8m)cos(theta)= .7 sin(theta)/cos(theta) =.7 tan(theta) = .7

this is sooo confusing! thank you soooooo much for helping me ! so would it be 9.8msin(theta) - (9.8m)cos(theta) * .7

the forces in the x direction are fgx and fst 9.8msin(theta) + (9.8m)cos(theta) * .7

okay so fgy = 9.8mcos(theta) and fgx = 9.8msintheta)

i thought the fg would break up into fg in and fg out and the fg in would match up fn. I am looking at a past problem we've done but we solved for acceleration

so would it be Fgx = 9.8m(cos theta) Fgy = 9.8m(sin theta)

are any of my equations in post #7 right?