Answer:
a) F = 4.9 10⁴ N, b) F₁ = 122.5 N
Explanation:
To solve this problem we use that the pressure is transmitted throughout the entire fluid, being the same for the same height
1) pressure is defined by the relation
P = F / A
to lift the weight of the truck the force of the piston must be equal to the weight of the truck
∑F = 0
F-W = 0
F = W = mg
F = 5000 9.8
F = 4.9 10⁴ N
the area of the pisto is
A = pi r²
A = pi d² / 4
A = pi 1 ^ 2/4
A = 0.7854 m²
pressure is
P = 4.9 104 / 0.7854
P = 3.85 104 Pa
2) Let's find a point with the same height on the two pistons, the pressure is the same
where subscript 1 is for the small piston and subscript 2 is for the large piston
F₁ = 
the force applied must be equal to the weight of the truck
F₁ = 
F₁ = (0.05 / 1) ² 5000 9.8
F₁ = 122.5 N
A. Move 2 m east and then 12 m east; displacement is 14 m east and the distance is 14 m
B. Move 10 m east and then 12 m west, the displacement is 2 m west and the distance is 22 m.
C. Move 8 m west and then 16 m east; the displacement is 8 m east and the distance is 24 m
D. Move 12 m west and then 8 m east; the displacement is 4 m and the distance is 20 m
<span>If an inductor is connected across an ac source and suppose the frequency of the source is doubled, then t</span>he inductive reactance of the inductor is also doubled. The inductive reactance (XL) is the t<span>he opposition to current flowing through a coil in an AC circuit, the </span>impedance measured in Ohms and can be calculated with the following formula:
XL=2*pi*f*L,
where f is the frequency. So, if the frequency is doubled than also the inductive reactance is doubled.
Answer:
V = 6.36 m³
Explanation:
For this exercise we will use fluid mechanics relations, starting with the continuity equation.
Let's write the flow equation
Q = v₁ A₁
The area of a circle is
A = π r²
Radius is half the diameter
A = π/4 d²
Q = v₁ π/4 d₁²
Q = π/ 4 15 0.03 2
Q = 0.0106 m3 / s
The volume of water in t = 10 min = 10 60 = 600 s
Q = V / t
V = Q t
V = 0.0106 600
V = 6.36 m³
