15.0 kg mass is displaced 3.00 m south and then 4.00 m west by a 10.0 N force. What is the total work done on the object? a. 25.0 J b. 30.0 J c. 40.0 J d. 70.0 J

Explanation:

3.00 m + 4.00 m = 7.00 m x 10.0 N = 70.0 J

found in the nucleus

has mass of one amu

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west