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:
a = 7.5 m / s²
Explanation:
For this exercise let's use Newton's second law, let's create a coordinate system with the x axis parallel to the plane and the y axis perpendicular to the plane
Y axis
N - W cos θ = 0
N = mg cos θ
X axis
W sin θ = m a
mg sin θ = m a
a = g sin θ
let's calculate
a = 9.8 cos 40
a = 7.5 m / s²
Answer:
f = 8 N
Explanation:
Data provided in the question
Radius of the pulley = r = 0.05 m
Moment of inertia = (I) = 0.2 kg.m^{2}
Angular acceleration = ∝ = 2 rad/sec
Based on the above information
As we know that
Torque is
And,
Torque is also
So,
We can say that
0.05f = 0.4
f = 8 N
We simply applied the above formulas
Explanation:
If the stones are unloaded from the boat, the weight of the boat will decrease. Therefore, the volume of the water displaced by the boat will also decrease. Due to this, the volume of the boat immersed in the water decreases. Hence, the level of the water around the boat will decrease.
Answer:
The angle of incident ray is 40°.
Explanation:
Given that the angle of incident and reflected ray are the same. In this question, we had given that both angles added up will gives you 80° so you have to divide it by 2 :
incident + reflected = 80°
Let incident = reflected = θ
θ + θ = 80°
2θ = 80°
θ = 80° ÷ 2
= 40°
