Tthe force exerted by the wind on the sails of a sailboat is 390 n north. the water exerts a force of 180 n east. if the boat (including its crew) has a mass of 270 kg, what are the magnitude and direction of its acceleration? serway, raymond a.. college physics (p. 114). brooks cole. kindle edition.

a= 1.59 m/s² : Magnitude  of the acceleration

β = 65.22°   (north of east) : Direction of the acceleration

Explanation:

Conceptual analysis

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Problem development

The acceleration is presented in the direction of the resultant force applied .

Calculation of the resultant forces (R)

R= 429.5 N

We apply the formula (1) to calculate the magnitude of the acceleration(a) :

∑F = m*a , m= 270 kg

R= m*a

429.5 =270*a

a= 1.59 m/s²

Calculation of the direction of the acceleration (β)

β = 65.22°   (north of east)

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.

the components of a vector are the parts of the vectors that are to each other.

b perpendicular