A hydraulic lift raises a 2000 kg automobile when a 500 N force is applied to the smaller piston. If the smaller piston has an a
1 answer:
Answer:
The cross-sectional area of the larger piston is 392cm ^{2}[/tex]
Explanation:
To find the solution, we apply the following equation:
Pascal's principle: F=P*A Formula (1)
F=Force applied to the piston
P: Pressure
A= Area of the piston
Nomenclature:
Fp= Force on the primary piston= 500N
W= weight of the car =m*g=2000kg*9.8m/s2= 19600N
Fs= Force on the secondary piston= W = 19600N
As= Area of the secondary piston=?
Pressure applied on one side is distributed to all liquid molecules since liquids are incompressible.
From equation (1)
P=F/A
Pp=Ps
You might be interested in
Answer:
the time it takes after impact for the puck is 2.18 seconds
Explanation:
initially given information
mass = 30 g = 0.03 kg
diameter = 100 mm = 0.1 m
thickness = 0.1 mm = 1 × m
dynamic viscosity = 1.75 × Ns/m²
temperature of air = 15°C
to determine
time needed for the puck to reduce its speed by 10%
solution
we note that velocity changes from 0 to v
assuming initial velocity = v
therefore final velocity = 0.9v
implying a change in velocity is du = v
and clearance dy = h
shear stress acting on the surface is expressed as
= µ
therefore
= µ ............1
substituting the values
= 1.75 × ×
= 0.175 v
the area between the air and puck is given by
Area =
area =
area = 7.85 × m²
thus, the force on the puck can be represented as
Force = × area
force = 0.175 v × 7.85 ×
force = 1.374 × v
now applying Newton's second law
force = mass × acceleration
- force =
- 1.374 × v =
solving for t =
the time needed after impact for the puck is 2.18 seconds