This question is incomplete, the complete question is;
A 1,040 N force is recorded on a hemispherical vane as it redirects a 2.5 cm- blade diameter water jet through a 180 angle.
Determine the velocity of the flowing water jet if the blade is assumed to be frictionless.
Answer: the velocity of the flowing water jet is 32.55 m/s assuming the blade is frictionless
Explanation:
Given that;
Force Ft = 1040 N
diameter d = 2.5 cm = 0.025 m
we know that; force acting on Hemispherical plate is;
Ft = 2δav²
where
a is area = π/4(0.025)²
δ is density of water = 1000 kg/m³
v is velocity = ?
now we substitute
1040 = 2 × 1000 × (π/4(0.025)²) × v²
1040 = 0.9817v²
v² = 1040 / 0.9817
v² = 1059.3867
v = √1059.3867
v = 32.5482 ≈ 32.55 m/s
Therefore the velocity of the flowing water jet is 32.55 m/s assuming the blade is frictionless
Técnicas y materiales de escultura
Modelado de arcilla. Clay tiene muchas ventajas para un artista. ...
Modelado de cera. El modelado con cera permite a un artista crear un diseño único. ...
Esculpir. ...
Talla de madera. ...
Talla de marfil. ...
Piedras semipreciosas y tallas de conchas. ...
Fundición de bronce. ...
Decoración.
Corrígeme si estoy equivocada.. :)
Answer:
25.46 MJ
Explanation:
continuity : m3 - m1 = -m<em>e</em>
<em>Energy Equation: m3u3 - m1u1 = -meue + 1Q3</em>
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<em>See the image attached (Well typed out format)</em>
Answer:
There are following type of hydraulic cylinders
1.Single acting
2.Double acting
3.Single acting spring loaded
4.Non differential type
5.Ram type
6.Piston type
7.Actuating type
8.Telescopic type
Answer:
In the steel: 815 kPa
In the aluminum: 270 kPa
Explanation:
The steel pipe will have a section of:
A1 = π/4 * (D^2 - d^2)
A1 = π/4 * (0.8^2 - 0.7^2) = 0.1178 m^2
The aluminum core:
A2 = π/4 * d^2
A2 = π/4 * 0.7^2 = 0.3848 m^2
The parts will have a certain stiffness:
k = E * A/l
We don't know their length, so we can consider this as stiffness per unit of length
k = E * A
For the steel pipe:
E = 210 GPa (for steel)
k1 = 210*10^9 * 0.1178 = 2.47*10^10 N
For the aluminum:
E = 70 GPa
k2 = 70*10^9 * 0.3848 = 2.69*10^10 N
Hooke's law:
Δd = f / k
Since we are using stiffness per unit of length we use stretching per unit of length:
ε = f / k
When the force is distributed between both materials will stretch the same length:
f = f1 + f2
f1 / k1 = f2/ k2
Replacing:
f1 = f - f2
(f - f2) / k1 = f2 / k2
f/k1 - f2/k1 = f2/k2
f/k1 = f2 * (1/k2 + 1/k1)
f2 = (f/k1) / (1/k2 + 1/k1)
f2 = (200000/2.47*10^10) / (1/2.69*10^10 + 1/2.47*10^10) = 104000 N = 104 KN
f1 = 200 - 104 = 96 kN
Then we calculate the stresses:
σ1 = f1/A1 = 96000 / 0.1178 = 815000 Pa = 815 kPa
σ2 = f2/A2 = 104000 / 0.3848 = 270000 Pa = 270 kPa
