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Questions (13)
The Wrong House by Guy de Maupassant Pere Varajou, formerly a horticulturist at Angers, but now retired from business, had
5 answers
58 views
the volume of fort peck dam is 96,000 X 1,000 m3. Suppose the state of Montana decides to increase the volume of the dam. After
2 answers
1,017 views
ok I'll be using a home work problem to ask the question. I'm not asking for help on the question I got the correct answer just
1 answer
592 views
I just need to know how to start this equation because it makes no sense
The position of a particle is given by r = (6.0 cos 3.0
0 answers
549 views
I still need help on my other question as well...
A runner hopes to complete the 10,000 m run in less than 30.0 min. AFter
3 answers
755 views
ok part (b)
A 4.0 kg block is stacked on top of a 12.0 kg block, whcih is accelerating along a horizontal table at a = 5.2 s^-2
4 answers
1,972 views
A 4.0 kg block is stacked on top of a 12.0 kg block, whcih is accelerating along a horizontal table at a = 5.2 s^-2 m. Let Mu k
8 answers
2,512 views
this one is kind of hard
A bicyclist can coast down a 7.0 degree hill at a steady 9.5 h^-1 km. If the drag force is proportional
13 answers
2,791 views
when resolving gravity into components
Fg (x component) = Fg sin theta when you get the units you get Newton times a degree how
1 answer
557 views
ok
Boxes are moved on a converyor blet from where they are filled to the packing station 10m away. The belt is initially
9 answers
1,032 views
ok
Boxes are moved on a converyor blet from where they are filled to the packing station 10m away. The belt is initially
0 answers
553 views
I'm trying to prove Kepler's equations
the proof is in my book but I don't understand it (T1/T2)^2 = (s1/s2)^3 ok first what
2 answers
634 views
ok were doing a lab
h t t p: / /i m g2 0 0. i m a ge s h a c k. u s / im g 2 0 0 / 30 3 8 /s k e t c h 2 p q g. j p g the timer
3 answers
723 views
Answers (48)
ya i realized that but I still have no idea how you get 2 - [(x-2)^2sgn(2-x)]/2 by hand...
Fg on object = (it's density)(it's volume)g Fb = (density of water)(volume of water underneath object pushing it up)gravity your suppose to divide the two gravity is suppose to cancel out and the volume but the volumes are two different objects why are
why are the volumes the same
ok so I got down to this x = sqrt( ((m1 + m2)(3.84 E 8))/m2 ) which gave me the wrong answer
m3 = mass of satalite (G m3 m2)/x^2 = (G m3 m1)/(3.84 E 8 m - x)^2 I can cancel out the m3 and the G right?
ok i get it see new question
Fg exerted by whatever it's orbiting onto moon = r^-2 G (mass of moon) M = to the gravitional force exerted by moon onto M = Mg ok then you can divide by M but why is it the force of gravity exerted by the moon onto whatever it's orbiting mass M just equal
ok net force = m a only force present force of gravity Fg = mg = r^-2 G m M were m is the mass of the moon and the M is the mass of what ever it's orbiting... correct? Fg = mg = r^-2 G m M divide both sides by the mass of the moon g = r^-2 G M so I guess I
Us = (rg)^-1 V^2 Us = (.12 m (9.80 kg^-1 N))^-1 (22.35 s^-1 m)^2 Us = 420 back of book says .34
50 mph is in the question in the book
which expression is correct from this 180 s(a t3) THIS ONE??? (180 s)a + (180 s)t3 or this ONE??? 180 s * a * t3
Is the answer to (c) correct by the way for the other problem I didn't know how fast it would be accelerating realtive to the larger block because the larger block would be accelerating more at 5.2 s^-1 m and the smaller block accelerates at 3.2 s^-1 m so
thanks for help on other question =]
ok for the applied force for (a) on the bottom block I got 83 N not sure if it's right... for the applied force in (b) wouldn't it be the same... I'm confused... Can you check my answer for (c)
can you check my answer for (c)
Sense this problem 32 the answer isn't in the back =[
also is the talbe frictionless or not I couldn't understand by the way it was worded
hmmmmmm interesting THANKS YOU
wouldn't it have a zero acceleration sense it's suppose to not be moving
Were did 4.5 s^-2 m come from???
Isn't net force x direction = -Ffr sense it's in the left direction then I would get a negative Mu???
i don't see how to apply Newtons second law to the top block there's a force pointing left the force of static friction is there a force opposing this force pointin in the right direction? How do I find the mininum coefficent of friction
ok for drag force c = v^-2 Fgx = (2.639 s^-1 m)^-2 95.5 N Fgx = mg sin theta
ok for b i got 565 N which was correct with book answer in back
ok I know this is right I redid it (a) 13.7 m^-1 kg
for (b) i got -.4172 N ;[
ok for (a) 13.7 s^-1 kg
part b ... a velocity of 25 h^-1 km is it constant velocity were there would be no acceleration if so then there would be no net force... so i'm lost
ok i'm stuck on part (b)
ok with correct sig figs (a) 13.7 m^-1 kg
ok for (a) 13.719 m^-1 kg
ok how is this for (a) 8.2 x 10^-14 m^-1 kg
ok fixing answer for (a)
wait I think i got it lol can you tell me if my answer for (a) is right???
oh wow... Thanks!!!
if you do this t=�ã(2S/a)=10/5.886=1.303 s. your doing the full distance 10 m instead of 5 m right??? so why would you double it
I did that
Acceleration for first half of trip... apply newtons second law... net force = m a = - F fr = - (Mu s Fn) were Mu s is coefficent of static friction take the inverse of m onto both sides a = m^-1 -(Mu s Fn) apply Newtons second law in the y direction net
oh and also what makes angle three equal to angle to if you extended the lines
what are these opposite angles that you are making reference to sorry it's been a while sense I took geometry
I agree and as you can see I wrote "...Vo^2 dosen't become Vo" not ""...Vo^2 dosen't become V"
I didn't right Vo^2 as V...
Thanks =]
no that was it thanks
what if the velocity is kept constant or is suppose to at least
they cancle each other out look (a = m2^-1 (m1 a) -g )a^-1 = 0 = m2^-1 m1 - g 0 = m2^-1 m1 - g
I got this a = m2^-1 (m1 a) -g i'm stuck from there
what's FBD how do i solve for F2
