Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?
Answer:
225 N
Explanation:
From Pascal's principle,
F/A = f/a ...................... Equation 1
Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.
Making f the subject of the equation,
f = F(a)/A ..................... Equation 2
Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².
Substituting into equation 2
f = 15000(3/200)
f = 225 N.
Hence the downward force that must be applied to small piston = 225 N
Answer:
W = -1844.513 J
Explanation:
GIVEN DATA:
mass of spider man is m 74 kg
vertical displacement if spider is 11 m
final displacement = 11 cos 60.6 = - 6.753 m
change in displacement is = -6.753 - (-11) = 4.25 m
gravity force act on spiderman is f = mg = 74 × 9.8 = 725.2 N
work done by gravity is 
where 180 is the angle between spiderman weight and displacement
W = -1844.513 J
Answer:
The value of the correct angle of banking for the road is 
°
Explanation:
Given data
Velocity (v) = 60 
Radius = 150 m
The velocity of the car in this case is given by
Put all the values in above formula we get
2.446
°
Therefore the value of the correct angle of banking for the road is °
<span>This is best understood with Newtons Third Law of Motion: for every action there is an equal and opposite reaction. That should allow you to see the answer.</span>
Answer:
F = 3600 [N]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of force must be equal to the product of mass by acceleration.
ΣF = m*a
where:
F = force [N]
m = mass = 2000 [kg]
a = acceleration = 1.8 [m/s^2]
Now replacing:
F = 2000*1.8
F = 3600 [N]
