Answer:
D. a person pulling a sled.
Explanation:
contact force only occurs when something directly comes in contact with another object.
a. is wrong because that is called magnetic force.
b. gravitational force
c. is an electrical force
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.
Answer:
the sum of all force being applied to an object.
Explanation:
Energy travels with sound waves.
The relationship between inductance and frequency can be clearly described using the following equation of inductive reactance:
Xl = 2*pi*f*L ; simplifying:
L = Xl / 2*pi*f
Therefore, as what we saw, inductance and frequency are inversely proportional. To add up, when inductance increases the frequency would decrease.