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:
the pic is blur,btw can u type sweetie
Answer:
Explanation:
Given that
As both charges are negative so there exist force of repulsion in direction as shown in figure.
Angle at which force F12 is acting is
Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction
