Recent content by lalalah

1. A fountain sends a stream of water straight up into the air to a maximum height of 4.23 m. The effective area of the pipe feeding the fountain is 5.38 x 10^-4 m^2. Neglecting air resistance and any viscous effects, determine how many gallons per minute are being used by the fountain. (1 gal =...

ahhhhhhh... so i found out that i should have punched in the answer as 2.4E21 instead of trying all the zeros! I'm an idiot. but thank you for all of the help! :)

ackkkk... so when i punched out the answer, i got 2.38 x 10^21 N, and naturally, it doesn't fit in the answer type-in box! does anyone know if i did anything wrong? maybe my part a wasn't as correct as i thought it was... i calculated part a using V_t = w * r, r = 33,000 Light...

awesomeee. thanks for your help!

i got part a, but i can't get part b! 1. A star has a mass of 2.96 x 10^30 kg and is moving in a circular orbit about the center of its galaxy. The radius of the orbit is 3.3 x 10^4 light-years (1 light-year = 9.5 x 10^15 m), and the angular speed of the star is 1.6 x 10-15 rad/s. (a) Determine...

wait let me try that again Fn = 2.37 (mv^2)/r = 2 mu *Fn (.00392*18.4^2)/r = 2*0.864*2.37 1.327/r = 4.095 r = .324 so it's the first answer divided by two! I'm starting to get the idea of how to use forces now... thank you so much for your help, i appreciate it!

ah i see. so in this case, i can say there are two friction forces because they do not cancel each other out? so it should be... (mv^2)/r = 2 mu *mg (.00392*18.4^2)/r = 2*0.864*.00392*9.8 1.327/r = 0.0664 r= 19.98 m ?

1. A stone has a mass of 3.92 g and is wedged into the tread of an automobile tire, as the drawing shows. The coefficient of static friction between the stone and each side of the tread channel is 0.864. When the tire surface is rotating at 18.4 m/s, the stone flies out of the tread. The...

so it's correct? yay! the weight of the car is negligable

I think I basically get the gist of this problem, but I want to know wjether i did everything correctly. So... 1. A car is towing a boat on a trailer. The driver starts from rest and accelerates to a velocity of +18.5 m/s in a time of 25 s. The combined mass of the boat and trailer is 562...

hoorayy~! and yet i would never have expected the number to come out so big... thank you for your help!

downward. so, would it be Fnet = -W + f_k = ma = -43.3 N? so... f_k = (96 kg)(9.8 m/s) -43.3 N and f_k = 897.5 N ? i hope its right :x

So, Fnet = -W + f_k = ma = 43.3 N? = -(mg) + f_k = ma = 43.3 N = -(96 kg)((9.8 m/s) + f_k = 43.3 N = f_k = 940.8 N does this look right? it doesn't seem right to me...

forces which act on the man: gravity, kinetic frictional force

oh sorry! i meant to type -0.4512 m/s^2! in that case, would it be correct of me to say that the net force and not the kinetic frictional force is 43.3 N ?