In a hydraulic lift, the radii of the pistons are 2.73 cm and 9.6 cm. A car weighing W 10.0 kN is to be lifted by the force of the large piston. (a) What force Fa must be applied to the small piston?N(b) When the small piston is pushed in by 8.9 cm, how far is the car lifted?mm(c) Find the mechanical advantage of the lift, which is the ratio W / Fa.
The battery light actually has little to do with the battery. It means your alternator isn't working. You'll still keep running until the battery dies.
a) From Pascal's law the pressure is constant and since p F/A we have Fa/Aa F2/A2 So Fa F2*Aa/A2 10.0x10^3N*π*(0.0273m)^2/(π*(0.096m)^2) 809N b) The volume of fluid moved is constant so Aa*ha A2*h2 so h2 ha*Aa/A2 8.9cm*π*(0.0273m)^2/(π*(0.096m)^2) 0.720cm c) MA W/Fa 10000/809 12.4
I bet a new battery fixes the problem.let a parts store charge the battery for a few hours. When it's charged, they'll test it to see if the volts drop and how fast.
I bet a new battery fixes the problem.let a parts store charge the battery for a few hours. When it's charged, they'll test it to see if the volts drop and how fast.
The battery light actually has little to do with the battery. It means your alternator isn't working. You'll still keep running until the battery dies.
a) From Pascal's law the pressure is constant and since p F/A we have Fa/Aa F2/A2 So Fa F2*Aa/A2 10.0x10^3N*π*(0.0273m)^2/(π*(0.096m)^2) 809N b) The volume of fluid moved is constant so Aa*ha A2*h2 so h2 ha*Aa/A2 8.9cm*π*(0.0273m)^2/(π*(0.096m)^2) 0.720cm c) MA W/Fa 10000/809 12.4
