Answers by visitors named: physics

Is the answer for the 1st question ...23.5 rad/ sec? Is the answer for the 2nd question...66 times? Is the answer for the 3rd question...3382.5 meters?

the other physics question

odd i never seen this formula before and we haven't goten to tension yet thank you

so its A

so Tf is 84.70?

can you please show me?

the confederates also thought that if they had a choice to be a united nation, they also had a choice to leave it.

11/5 is an improper fraction, once you divide it, it will become a mixed number like MS.Sue said, 2 1/5

wow, it sounds really good but i think there might be a few spelling or grammatical errors, i found one:like 'seamed' which should be 'seemed'

nevermind i think its A can you check that?

thanks!

then it's A?

so, how would i write it out? = -0.544 - -3.40 ????????

ok, thanks...i just want few other people to reply to this post...just to be sure i agree with you but int the one "which of the following statements about electromagenetic..." i'm not sure between B and D thanks again

but i checked and it does point in the opposite direction the charges are differnect signs..... i'm confused!

is it .24 amps?

ok, thanks

y^2 + 6y -12 y -72 simplify

I was taking upwards in the y direction positve and downwards in the y dierction negetive my bad

I think you mean the limits

I took algebra two and there not really verticle asymptotes threr limits you have been lied to which you should be used to if you are going to a public high school All I remeber is you try to get them to cancel out and get removable discontinuties or something this was a long time ago so I can't really help

thank you so much

I don't see how this relationship works Fn sin theta = r^-1 m v^2 or more like why it works??? my text goes on to prove tan theta = (rg)^-1 v^2 which I understand how to rearange I just don't get where the first relationship comes from Fn sin theta = r^-1 m v^2 i get its the horizontal component of the normal force but I don't see why we can set it equal to the radial acceleration

also I was always told to use this equation when the velocity is small through the medium Fd = -bv were this is just a proportion also that other formula that is suppose to be used when there is a high velocity Fd = 2^-1 p v^2 A Cd v Ok what is exactly meant by high and low velocity and how do you derive the equation for a high velocity all I've been told was the equation never taught how it was derived and don't know were to start

I just would like to make sure of something an object experiances a drag force when it travels through a medium like air or water... so dosen't every thing experience a drag force that is in motion because its moving through air?? So like if a person was walking do they experiance a drag force? I would think so... just making sure

your image didn't work

link didn't got to image

hey wow thanks!!! I acutally understood just kind of werid because our book hasn't mentioned energy yet but I know about it from just regualr old physics I'm taking ap this year and doing the first chapters in order and it hasn't mentioned anything about energy yet just told me it was twice but not why Thanks =]

not for me =[ look... h t t p : / /i m g1 3 2 . i ma g e s h a c k. u s / i m g1 3 2 / 9 1 3 7 /c a p tu r e t q i. j p g could you try image shack upload (link below) I don't know why it wouldn't let me see it i captured the screen when i clicked on the link it's the link above h t t p : / / i m a g e s h a c k . u s / =]

when i click on your link it bring me here h t t p : / / ww w . a n g e l f i r e . l yc o s . c o m / d o c / im ag e s / r l o a d b l oc k . j p g i noticed a different adress then the link

what's fw and Fp

ohh ok um Fw is the gravity and Fp is the x component and Fn is the y component??? how come the x component is drawn in a weired location but we got the triangle now thanks!!!

How to prove this? Similar triangles. Notice the upper angle in that photo is a right angle (ramp to Fn), yes and i agree right anlge in which the angle of Fn to the vertical is an angle complement to Fp and the vertical. ok so your saying that... angle Fn Fw is complement to angle fp and fw because triangle adds up to 180 degrees ok understand Well, the ramp angle is also a complement to that angle. huh so angle of the ramp is complemtary to angle Fw Fp... ok i can visually see that but how do we now these two angles are complementary??? Therefore, ramp angle = angle Fn to vertical. I agree... now snese both angles are completary to this other angle then they must be equal THANKS!!!

ha ha i copied the lnik instead of clicking on it i got it thanks

so there is no static friction when a block is just sitting there and not moving with no applied force?

Net Force Radial = m a radial = F radial = Ffr ma radial = Us Fn ma radial = Us m g cancel out mass a radial = Us g a radial = r^-1 v^2 r^-1 v^2 = Us g Us = (rg)^-1 V^2 don\'t see what I did wrong

If this question is asking for the normal force i got 100 n but i\'m not sure that is what it is asking for

this is from that white book that i'm also using... the odd answers in the book are in the back in which this problem is one I belive

I get 131.18 degrees and its wrong. I don't know what I am doing wrong.

I typed in the answer for the direction (85 and 175) and they are both wrong according to the problem. I want to try 4.8 degrees since it says east of south, but I don't know though.

density=mass/volume volume-mass/density

The problem has 55m/s... the instructor changed numbers in the original problem and I guess he wanted to change that number to 55. I thought it was strange too at first. So for conservation of momentum, would i use mhvh=mcvc where h is for human and c is for car?

The vertical acceleration is 9.8 m/s^2 because of gravity. So in the negative y direction.

I think it's saying only gravity is affecting the particle on the y-axis.

A 59-kg housepainter stands on a 13-kg aluminum platform. The platform is attached to a rope that passes through an overhead pulley, which allows the painter to raise herself and the platform. (Ignore the mass of the pulley and any friction in the pulley.)

A. Because on a fricitionless horizontal surface any finite mass will accelerate if any finite net force is applied in any direction that is not perpendicular to the said surface.

Boring physics

exam question

i will not sure. This answer is write or wrong because this some is very difficult.sorry

do yourself

techari gaide?

review questions

Unit 3 end of unit questions

did this much already

nvm got the answers. Got 3.8m/s^2 and 5.6m/s^2 if anyones interested.

nevermindddd

neverminddd

Radiation occurs even in a vacuum, as matter can give off energy in the electromagnetic spectrum, depending on its form and temperature.

dont need answer anymore

thank you!!!

oh well LOL

hopefully got the right answer

A girl walk 5m, west, than walk 15m east. what is the displacement