Answer:
50J
Explanation:
At the top you have(A)
KE_a = O
PE_a = 100J
KE + PE = 100J
At the bottom you have (C)
KE_c= 100J
PE_c=0J
KE+PE = 100J
At point C:
You are at half the height.
We know that at H, PE =100J
PE_c = mgH
At C,
PE_c= mg (H/2) *at half the height
*m and g stay the same
Intuitively, the higher you are, the more potential energy you have.
If you decrease the height by a half, your PE will also decrease
At A:
PE_a / (mg) = H
At B:
PE_b / (mg) = H/2
to also get H on the right hand side, multiply by 2
2 (PE_b/ (mg))= H
2PE_b / (mg) = H
Ok, now that we have set up 2 equations (where H is isolated), find PE at B
AT A = AT B *This way you are saying that H = H (you compare both equations)
PE_a / (mg) = 2x PE_b / (mg)
*mg are the same for both cancel them (you can do that because of the = sign)
PE_a = 2PE_b
We know that PE_a = 100J
100J/2 = PE_b
PE at b = 50J
**FIND KE at b
We know that
KE_b + PE_b is always 100J
100J = 50J + KE_b
KE_b = 50J
Answer:
400W
Solution :
power (w) = work done (J) ÷ time(s)
x= 24000J ÷ 60s = 400W
Some elements that have a symbol entirely different from the spelling are..
1.Fe. Iron
2.Na.Sodium
3.K. Pottasium
4.Ag.Silver
5.Sn.Tin
6.Sb. Antimony
7,Pb.Lead
Answer:
4.465 m
Explanation:
Taken the beam to be uniform, then the center of gravity will act at the mid-point of the beam. The mid point = 11 / 2 = 5.5 m
since the boy can walk to the end of the beam without falling,
then the torque ( moment of the force) by beam = torque by students at the end of the beam
let the distance of the boy = x
220 ( 5.5 - x ) = 51 × x
1210 - 220 x = 51 x
1210 = 51 x + 220 x
1210 = 271 x
x = 1210 / 271 = 4.465 m