Answer:
(A). The work done by friction in crossing the patch is -637.98 J.
(B). The speed of skier is 10.57 m/s.
Explanation:
Given that,
Mass of skier = 62 kg
Speed = 6.5 m/s
Length = 3.50 m
Coefficient kinetic friction = 0.30
Height = 2.5 m
(A) we need to calculate the work done by friction in crossing the patch
Using formula of work done

Put the value into the formula


The work done by friction in crossing the patch is -637.98 J.
(B) we need to calculate the speed of skier
Using conservation of energy


Final potential energy is zero
So, 

Put the value into the formula



The speed of skier is 10.57 m/s.
Hence, (A).The work done by friction in crossing the patch is -637.98 J.
(B).The speed of skier is 10.57 m/s.
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Answer:

Explanation:
Given:
- Three identical charges q.
- Two charges on x - axis separated by distance a about origin
- One on y-axis
- All three charges are vertices
Find:
- Find an expression for the electric field at points on the y-axis above the uppermost charge.
- Show that the working reduces to point charge when y >> a.
Solution
- Take a variable distance y above the top most charge.
- Then compute the distance from charges on the axis to the variable distance y:

- Then compute the angle that Force makes with the y axis:
cos(Q) = sqrt(3)*a / 2*r
- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:
F_1,2 = 2*F_x*cos(Q)
- The total net force would be:
F_net = F_1,2 + kq / y^2
- Hence,

- Now for the limit y >>a:

- Insert limit i.e a/y = 0

Hence the Electric Field is off a point charge of magnitude 3q.
If Kai takes the burger off of the grill and puts cheese on it, it melts because the burger still is hot from being on the grill. The heat in and on the burger doesn’t go away immediately, so that is how and why the cheese melts.