In a hydraulic garage lift, the smallpiston has a radius of 5.0 cm and the large piston has a radius of15 cm. What force must be applied on the small piston in order tolift a car weighing 20,000 N on the large piston? a. 6.7 ×103 N
b. 5.0 ×103 N
c. 2.9 ×103 N
d. 2.2 ×103 N

Option  d) 2.2 × 10³ N is the correct answer

Explanation:

Given that;

radius of A₁ = 5.0

large piston has a radius of A₀ = 15 cm

Weight F₀ = 20,000 N

Force that must be applied to lift the weight F₁ = ?

Now we know that from Pascal Law

ΔP ⇒ F₁/A₁ = F₀/A₀

F₁ =  (F₀/A₀)A₁

where A = πr²

so we substitute

F₁ =  (20,000 / π(15²)) × π(5²)

= (20,000 / 706.8583) × 78.5398

= 2222.22 ≈ 2.2 × 10³ N

Therefore Option  d) 2.2 × 10³ N is the correct answer.

i dont have a visual in front of me, but i can you reason to try to get the right one. the graph should look like this, 0 to 6 there is a spike, 12 to 14 there should be a very low point. slight increase in the graph at 12 to 14, and flat 14 to 18

explanation: