Charlie runs an auto repair and mechanic shop. A hot rod has come in with mass 1500 kg requiring a repair. He needs to raise the car using a hydraulic lift on a piston (Piston A) having a radius of 25 cm. What force needs to be applied to the opposing 1.0 cm radius piston (Piston B) to accomplish this? Ignore the weight of the pistons.

23.52 N

Explanation:

Using the Hydraulic Press formula

since the the liquid in an hydraulic press cannot be compressed

P₁ ( pressure at the piston 1) = P₂ ( pressure at piston 2)

P₁ = F₁ / A₁ where F₁ represent the weight of the car and F₂ represent the force applied at the 1 cm end and A₁ and A₂ are the area of the piston respectively

P₂ = F₂ / A₂

F₁ / A₁ = F₂ / A₂

(F₁/A₁) × A₂ = F₂

F₁ = 1500 kg × 9.8 m/s² = 14700 N

A₁ = πR² = π ( 0.25)²  where R = 25 cm = 0.25 m for the larger piston

A₂ = πr² = π ( 0.01)²   where r = 1 cm = 0.01 m

(F₁/A₁) × A₂ = F₂

F₂ = ( 14700 N / ( π 0.0625)) ×π (0.01)² = 23.52 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

answergravity;