To solve this problem we will use the concepts related to the resulting Vector Force product of two components, that is,

If we take the Force of 50 N as the force in the X direction and the Force of 40 N in the Y direction we will have to:



Finally, since Newton's second law, acceleration can be determined as




Therefore the resultant magnitude of the acceleration of the object is 
 
        
             
        
        
        
Answer:
μ = 0.385
Explanation:
Given that,
The mass of the student, m = 69 Kg
The horizontal force applied, F = 260 N
The normal force acting on the body, weight = mg = 69g  N
                                                                                     = 676.2 N
The coefficient of friction acting on a body is equal to the force acting on the body to the normal force acting on the body due to gravitation.
The formula for coefficient of friction,
                                     μ =  F / N
 Substituting the values in the above equation,
                                      μ = 260 N / 676.2 N
                                         = 0.385
Hence, the  coefficient of friction, μ = 0.385
 
        
             
        
        
        
Answer:
A) ΔU = 3.9 × 10^(10) J
B) v = 8420.75 m/s 
Explanation:
We are given;
Potential Difference; V = 1.3 × 10^(9) V
Charge; Q = 30 C
A) Formula for change in energy of transferred charge is given as;
ΔU = QV
Plugging in the relevant values gives;
ΔU = 30 × 1.3 × 10^(9)
ΔU = 3.9 × 10^(10) J
B) We are told that this energy gotten above is used to accelerate a 1100 kg car from rest.
This means that the initial potential energy will be equal to the final kinetic energy since all the potential energy will be converted to kinetic energy.
Thus;
P.E = K.E
ΔU = ½mv²
Where v is final velocity. 
Plugging in the relevant values;
3.9 × 10^(10) = ½ × 1100 × v²
v² = [7.8 × 10^(8)]/11
v² = 70909090.9090909
v = √70909090.9090909
v = 8420.75 m/s
 
        
             
        
        
        
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the weight of the phone? 1 kg of mass is about 2.2 lb. is no. 3 which is "<span>No; it weighs about 45 lbs."</span>
        
             
        
        
        
There is a horizontal abduction and horizontal adduction in the
transverse plane that is imaginary bisector that divides the body into two halves
upper or top and lower or bottom halves.
<span>Horizontal abduction involves the movement of the arm or thigh from
a medial or anterior position to a lateral position and in horizontal
adduction, the movement is reverse that is from lateral position to anterior
position.</span>