Recent content by Apple123

Ahh i get it. Ok I am off on that part, but it works since the object is being pull downed towards the Earth so it would add to the tension. Anywhere besides straight down though, my calculations would be totally off. Except at the exact top of the circle, where the tension would be F(c) minus...

I think part 2 of the problem is right, since it doesn't deal with the tension. I am still confused about part 3 though also. According to my book: At the bottom of a vertical circle: F(centripetal)=F(normal)-mg I assume that F(normal)=T (thats what i always thought, but I am not sure)...

ok first i solved the x equation to be: T(a)=T(b)(cosB/CosA) Then i substituted it into the y equation. I moved mg to the right side and factored out T(b) from the numbers. Then i solved for what was left on the right side (the sines and cosines) to receive T(b)=42.4785N Does this sound...

-------------------------------------------------------------------------------- A 5.0 kilogram monkey hangs initially at rest from two vines, A and B. Each vine is 10 meters in length with negligible mass. Vine A is on the left with an angle of 30 degrees from the horizontal from the...

Show me some work, and then ill help you. Or atleast convince yourself of the concept. I'd be happy to help, but i don't want to do it for you.

What a notice is that angles of A and B are complements, so if i messed up using sine and cosine i would have done the opposite vine. I can't figure out if I'm using the right equations and solving them the right way.

torque=(mag of force)(radius of lever arm) you already have the radius, so just find the force now What do you have to find the force? ill help you from there

A 5.0 kilogram monkey hangs initially at rest from two vines, A and B. Each vine is 10 meters in length with negligible mass. Vine A is on the left with an angle of 30 degrees from the horizontal from the monkey and B is at 60 degrees from the horizontal of the angle next to the monkey. 1...

Here is the problem: A 10.0-kg crate is pulled up a rought 20 degree incline by a 100-N force parallel to the incline. The initial speed of the crate is 1.5 m/s, and the coefficient of kinetic friction is 0.40. The crate is pulled a distance of 5.00m. Determine the work done by the: 1...

After i did that equations, and the plugged them into the third equation, I got the value of 2.15x10^-14, does this seem about right or did i get off somewhere?

Ohhh, so would u first solve Oy for T, then sub T into Ox and solve for F then take the absolute value? Then when u solve for the next equation, are the two vacuum permitivites both 8.85.10^12? I've never heard of this concept before.

ohhh ok i get it. Now to solve for F electric can u just use F=ma? And then for the two vacuum permitivities, are they the same number, or are they different?

In the k equation, what is the symbol that follows the 4pi?

Ok i understand every part to the problem except the 4Pi and the things after it up until the parentheses. What do those mean, and where did they come from.

I found this equation online, does it work? http://homepage.smc.edu/physsci/dept/Physics/P22Hwk/EField.htm Go there, its on number 5. Its the equation where q=2Lsin... If so, how is it derived?