QUESTION

This page lists questions and answers that were posted by visitors named QUESTION.

Questions

The following questions were asked by visitors named QUESTION.

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 in the picture makes a mark on the tape at a set interval... my question is well what is actually is happening is that I'm pull...

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 allows us to rewrite that formula as so??? (s1^3/T1^2)=(s2^3/T2^2) my book goes about the proof as m1 ((4 pi^2 r1)/T1^2) which I g...

when resolving gravity into components Fg (x component) = Fg sin theta when you get the units you get Newton times a degree how come we just ignore the degree for example if it was 9 N you wouldnt say 9 N (degrees) why not? How do the degrees just magical...

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 to the square fo the speed v, so that Fd = cv^2, calculate (a) the value fo the constant c and (b) the average force that must...

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 = Mu s = Mu. (a) What mininum coefficent of friction Mu between the two blocks will prevent the 4.0 kg block from sliding off?...

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 running at constant speed exactly 27.0 min there are still 1100 m to go. The runner must then accelerate at .20 s^-2 m for how many...

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 need to understand this concept better. Calculate the acceelration due to gravity on the Moon. The moon's radius is about 1.74...

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 the improvements, fort peck will hold 10 times as many cubic meters will fort peck hold after the improvemts?

The Wrong House by Guy de Maupassant Pere Varajou, formerly a horticulturist at Angers, but now retired from business, had closed his purse strings to his scapegrace son and had hardly seen him for two years. His daughter had married Padoie, a former trea...

Answers

The following answers were posted by visitors named QUESTION.

what's FBD how do i solve for F2

I got this a = m2^-1 (m1 a) -g i'm stuck from there

they cancle each other out look (a = m2^-1 (m1 a) -g )a^-1 = 0 = m2^-1 m1 - g 0 = m2^-1 m1 - g

what if the velocity is kept constant or is suppose to at least

no that was it thanks

I didn't right Vo^2 as V...

I agree and as you can see I wrote "...Vo^2 dosen't become Vo" not ""...Vo^2 dosen't become V"

what are these opposite angles that you are making reference to sorry it's been a while sense I took geometry

oh and also what makes angle three equal to angle to if you extended the lines

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...

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

oh wow... Thanks!!!

wait I think i got it lol can you tell me if my answer for (a) is right???

ok fixing answer for (a)

ok how is this for (a) 8.2 x 10^-14 m^-1 kg

ok for (a) 13.719 m^-1 kg

ok with correct sig figs (a) 13.7 m^-1 kg

ok i'm stuck on part (b)

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 for (a) 13.7 s^-1 kg

for (b) i got -.4172 N ;[

ok I know this is right I redid it (a) 13.7 m^-1 kg

ok for b i got 565 N which was correct with book answer in back

wouldn't it have a zero acceleration sense it's suppose to not be moving

ok for drag force c = v^-2 Fgx = (2.639 s^-1 m)^-2 95.5 N Fgx = mg sin theta

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

Isn't net force x direction = -Ffr sense it's in the left direction then I would get a negative Mu???

Were did 4.5 s^-2 m come from???

hmmmmmm interesting THANKS YOU

also is the talbe frictionless or not I couldn't understand by the way it was worded

Sense this problem 32 the answer isn't in the back =[

can you check my answer for (c)

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)

thanks for help on other question =]

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...

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

50 mph is in the question in the book

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

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...

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 equa...

ok i get it see new question

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?

why are the volumes the same

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 th...

ya i realized that but I still have no idea how you get 2 - [(x-2)^2sgn(2-x)]/2 by hand...