Answer:
Web Browser
Explanation:
Because you dont use a messaging app or presentation software to look up stuff its common knowledge
Answer:
Q= 4.6 × 10⁻³ m³/s
actual velocity will be equal to 8.39 m/s
Explanation:
density of fluid = 900 kg/m³
d₁ = 0.025 m
d₂ = 0.05 m
Δ P = -40 k N/m²
C v = 0.89
using energy equation
under ideal condition v₁² = 0
v₂² = 88.88
v₂ = 9.43 m/s
hence discharge at downstream will be
Q = Av
Q =
Q =
Q= 4.6 × 10⁻³ m³/s
we know that
hence , actual velocity will be equal to 8.39 m/s
Answer:
The part of the system that is considered the resistance force is;
B
Explanation:
The simple machine is a system of pulley that has two pulleys
The effort, which is the input force at A gives the value of the tension at C and D which are used to lift the load B
Therefore, we have;
A = C = D
B = C + D = C + C = 2·C
∴ C = B/2
We have;
C = B/2 = A
Therefore, with the pulley only a force, A equivalent to half the weight, B, of the load is required to lift the load, B
The resistance force is the constant force in the system that that requires an input force to overcome in order for work to be done
It is the force acting to oppose the sum of the other forces system, such as a force acting in opposition to an input force
Therefore, the resistance force is the load force, B, for which the input force, A, is required in order for the load to be lifted.
Answer:
the minimum shaft diameter is 35.026 mm
the maximum shaft diameter is 35.042mm
Explanation:
Given data;
D-maximum = 35.020mm and d-minimum = 35.000mm
we have to go through Tables "Descriptions of preferred Fits using the Basic Hole System" so from the table, locational interference fits H7/p6
so From table, Selection of International Trade Grades metric series
the grade tolerance are;
ΔD = IT7(0.025 mm)
Δd = IT6(0.016 mm)
Also from Table "Fundamental Deviations for Shafts" metric series
Sf = 0.026
so
D-maximum
Dmax = d + Sf + Δd
we substitute
Dmax = 35 + 0.026 + 0.016
Dmax = 35.042 mm
therefore the maximum diameter of shaft is 35.042mm
d-minimum
Dmin = d + Sf
Dmin = 35 + 0.026
Dmin = 35.026 mm
therefore the minimum diameter of shaft is 35.026 mm