d=n^2 - 3n / 2
where d = number of diagonals
and n = the number of sides of the polygon.
A polygon has 119 diagonals, how many sides does it have?
119 = n^2 - 3n / 2
I am not sure how to do this? can someone please help?
sorry this isn't physics related
I know I need to use...
d=n^2 - 3n / 2
where d = number of diagonals
and n = the number of sides of the polygon.
A polygon has 119 diagonals, how many sides does it have?
119 = n^2 - 3n / 2
I am not sure how to do this? can someone please help?
sorry this isn't physics related
[SOLVED] Current question
Hey, I have been working on this electric circuit section in class for a couple of days. I was just wondering if I am on the right track with this question seems how my teacher won't give me a very straight answer.
Six volts is applied to a 2 ohm resistor in series...
is mgh=1/2mv^2=sqrt(2*20*9.8) a shortcut or is that the way that I should setup the problem from the start? What is the 2 for in the equation. I understand that 20 = h and 9.8 = g
A 0.25 kg pine cone falls from a branch 20 m above the ground.
A) With what speed would it hit the ground if air resistance could be ignored?
m= 0.25 kg
g= 9.8 m/s^2
d= 20 m
Ep= (0.25kg)(9.8m/s^2)(20m)
= 49 J
Ek= 1/2mv^2
49J = 1/2(0.25kg)(v^2)
2(49 J = (0.5kg)(0.5 v^2))
98 J =...
An object falls from rest for a distance of 25 m. If there was no air friction, how fast was it going at that distance?
I'm not sure but I would imagine v=(25m)(9.8m/s^2)= 245m/s ? which seems quite high to me so I know I am not calculating it correctly.
yeah so it is v(final)=29.4 m/s + a(t)
but doesn't an object come back down at the same velocity that it went up?
that is why I thought:
-29.4 m/s = 29.4 m/s + a(t)