Posts by emath
Total # Posts: 8
What is the name of NH4CNO? is it ammonium something? Thank you
3sqrt(45) simplifies to: 3sqrt(9 * 5) 3*sqrt(9)*sqrt(5) 3*3*sqrt(5) 9sqrt(5) which equals about 20.1246
and for part a I realized that it is 0 because the particle is at the origin and I have to find the angular momentum of the particle about the origin: a = vi x= r r = 0
My apologies The "following locations" a) at the origin b)at the highest point of its trajectory c)just before it hits the ground the dimensions of angular momentum: ? kg*m^(2)/s the picture is simply a parabola, a projectile launched with some angle theta with ...
A particle of mass m is shot with an initial velocity vector v i and makes an angle θ with the horizontal, as shown below. The particle moves in the gravitational field of the Earth. (makes a shape of a parabola)Find the angular momentum of the particle about the origin ...
A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.24 m/s on a horizontal section of a track. It rolls around the inside of a vertical circular loop 90.0 cm in diameter and finally leaves the track at a point 19.0 cm below the horizontal...
MATH 4TH GRADE
5% converts to 0.05 in decimals of means multiplication so... 0.05 x 600= 30
integrate (sinx)^(5)(cosx)^(12)from 0 to (pi/2) I'm having problems just even integrating it... I tried breaking it up (sin^(2)x)^(2)(cos^(2)x)^(6)(sinx)dx then doing u-sub with u=cosx and du= -sinx but then I get stuck... the answer is 8/3315 just don't know how to ...