The large piston in a hydraulic lift has an area of 2m^2. What force must be applied to the small piston with an area of .2m^2 in order to raise a car that weighs 14,700 Newtons

147,000N

Step-by-step explanation:

A1= 2m^2

A2= 0.2m^2

F2= 14,700N

Required

F1, the applied force

Applying the formula

F1/A1= F2/A2

substute

F1/2=14700/0.2

2*14700= F1*0.2

29400= F1*0.2

F1= 29400/0.2

F1=147,000N

Hence, the applied force is 147,000N

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v