A10 kg lawn mower is pushed along by a force applied at a 20 degree angle to the horizontal. if a the mower accelerates at 3.4 m/s^2, what is this force?

m = mass of lawn mower = 10 kg

a = acceleration along X-direction = 3.4 m/s²

let the applied force be "F"

from the force diagram , we see that only "F Cos20 " is acting along the horizontal direction .

hence F' = net force along x-direction = F Cos20

net force along x-direction is given as

F' = ma

F Cos20 = ma

inserting the values

F Cos20 = 10 x 3.4

F Cos20 = 34

F = 36.2 N

hence the force applied is 36.2 N  

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j