Askier with a 65kg mass skies down a 30 degree incline hill. the coefficient of friction is 0.1. a. draw a free body diagram. b. determine the normal force. c. determine the acceleration of the skier down the hill.

Part a.
Refer to the diagram shown below.

m = 65 kg, the mass of the skier
θ = 30°, the angle of the incline
μ = 0.1, the coefficient of friction
W = mg , the weight of the skier
N =  mg cosθ, the normal reaction
F = mg sinθ, the component of the weight acting down the hill
R = μN, the frictional force resisting motion down the plane
a = the acceleration down the plane.

The weight is
W = mg  = (65 kg)*(9.8 m/s)  = 637 N

Part b.
The normal force is
N = (637 N)*cos(30) =  551.658 N

The frictional force is
R = 0.1(551.658 N) = 55.1658 N

Part c.
Calculate the component of the weight acting down the hill.
F = mg sin(30) = (637 N)*sin(30) = 318.5 N

The equation of motion is
ma = F - R
(65 kg)*(a m/s²) = 318.5 - 55.1658 N
a = 263.3342/65 = 4.05 m/s²

Answers:
N = 551.7 N
a = 4.05 m/s²

So what is the question

It won't lose or gain electrons if it's a covalent bond it shares the electrons with another atom or atoms. if it is an ionic bond it will lose the 3 electrons on the outer shell and become a positively charged ion.  credit goes to @sunshinebaby of answers.yahoo.com.