Answer:
a) the velocity of the water at the exit for the maximum lift condition is 3.62m/s
b) the discharge through the system for maximum lift condition is 0.00326m³/s
c) the maximum load the suction can support is 8.85KN
Explanation:
the explanation has been attached below , check for better understanding of the answers given above.
Answer:
The block will float with its axis vertical.
Explanation:
For it to float, the upward force on the cylindrical block must be equal to the weight of an equal volume of water. Also, this upward force must be greater than or equal to the weight of the cylindrical block for it to float.
So, weight of cylindrical block, W = specific weight × volume
specific weight = 7500 N/m³
volume = πd²h/4 where d = diameter of block and h = height of block
volume = π(1 m)² × 1 m/4 = π/4 m³ = 0.7854 m³
W = 7500 N/m³ × 0.7854 m³ = 5890.5 N
Since the density of water = 1000 kg/m³, its specific weight W' = 1000 kg/m³ × 9.8 m/s² = 9800 N/m³
Since the volume of the cylinder = volume of water displaced, the weight of water displaced W' = upward force = specific weight of water × volume of water displaced = 9800 N/m³ × 0.7854 m³ = 7696.92 N
Since W' = 7696.92 N > W = 5890.5 N, <u>the block will float with its axis vertical since the upward force is greater than the weight of the cylindrical block.</u>
Answer:
The torque on the section is 563.373 Nm
Explanation:
The torque in the section can be calculated through torsion formula. The torsion formula is given as follows:
Torque = T = τJ/ r
where,
r = maximum radius = 5.5 cm/2 = 2.75 cm = 2.75 x 10^-2 m
τ = maximum shear stress = 22 MPa = 22 x 10^6 Pa
J = Polar moment of inertial = π/2 (ro^4 - ri^4)
J = π/2 (0.0275^4 - 0.01875^4) m^4 = 7.042 x 10^-7 m^4
Therefore:
T = (7.042 x 10^-7 m^4)(22 x 10^6 Pa)/(2.75 x 10^-2 m)
<u>T = 563.373 Nm</u>
Answer:
- |z*| = r , ∠(z*) = -∅
- |z²| = r² , ∠(z²) = 2∅
- |jz| = r , ∠(jz) = ∅
- |zz*| = r² , ∠(zz*) = 0
- |z/z*| = 1 , ∠(z/z*) = 2∅
- |1/z| = r ⁻¹ , ∠(1/z) = -∅
Explanation:
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1.) The * mean you take the conjugate of the value. This means you change the sign of the imaginary part, so if it's positive, turn it negative and vice versa.
2.) The magnitude with an exponent can have the exponent moved outside the magnitude. |zⁿ| = |z|ⁿ
The angle multiplies with its exponent instead. ∠zⁿ = n∠z
3.) This part is just testing if you can convert the number using the eulers formula and convert back.
The magnitude could be found using the distance formula. √(R² + I²)
The angle could be found using tan⁻¹(Imaginary/Real).
4.) Magnitude of a product could be split up. |zv| = |z|·|v|
Angle of a product could be splitted up and added. ∠(zv) = ∠z + ∠v
5.) Simplify it first using some algebra and use the euler's identity to identify the magnitude and angle. It takes in a form like this:
A is your magnitude, ∅ is your angle.
6.) Same rule as part 2