Recent content by Jujubee37

I have attempted to plug in grams, kiliograms, and even pounds but it was wrong.

Okay thank you I got it I think. 9 W

would that be FD=1/2mv^2 ?

I started of by trying to find the work for I got stuck because I did not know how to solve for the Force. I solved for the distance by doing 3=d/5s which gave me 15m. but then I couldn't figure out where to go next in the problem because I don't know how to solve for (f) so an explanation would...

Yes I am aware now, my way got me the correct answer but yes plugging in 164 would be quicker.

The answer I got was 104.79 N. I have practiced more problems similar to this one so I understand it now.

ohhh okay that explained a lot. Thank you I got the problem right.

Yes okay thanks. I have two formulas for that. v=2pir/t and f=mv^2/r. If I used those equations how would I apply those to the problem because I was not given numbers to plug in.

I was thinking it would 3 or 9 times the tension rather than 1/3 or 1/9 but that is just a guess. an explanation would help very much!

i got the answer already. 16.73 is the mass. I had to solve for it

Do what? I really do not understand where I am supposed to start. I have no understanding of this problem. This is what the hint was for this problem. The net vertical force here is zero. The normal force is one of the vertical forces. Others are weight and part of the pulling force.

I still do not understand what I have to do. Solve for the horizontal? or what?

Lon capa the website. I don't understand this so I can't answer your questions.

I wouldn't know. All I know is that the vertical net force is zero and the normal force is one of the vertical forces. I worked this out two ways. one with sin and the other with cos because those are the only equations that I have involving an angle.

I did. (16.73)(9.8)-51.25(sin)(49.1)= 125.21 N