Answer:
The answer is VN =37.416 m/s
Explanation:
Recall that:
Pressure (atmospheric) = 100 kPa
So. we solve for the maximum velocity (m/s) to which water can be accelerated by the nozzles
Now,
Pabs =Patm + Pgauge = 800 KN/m²
Thus
PT/9.81 + VT²/2g =PN/9.81 + VN²/2g
Here
Acceleration due to gravity = 9.81 m/s
800/9.81 + 0
= 100/9.81 + VN²/19.62
Here,
9.81 * 2= 19.62
Thus,
VN²/19.62 = 700/9.81
So,
VN² =1400
VN =37.416 m/s
Note: (800 - 100) = 700
Answer:
40π
Explanation:
First, find the limits (intersections).
5x² = 30x − 10x²
15x² − 30x = 0
x² − 2x = 0
x (x − 2) = 0
x = 0 or 2
Within this interval, 30x − 10x² is greater than 5x².
Dividing the volume into cylindrical shells, the volume of each shell is:
dV = 2π r h t
dV = 2π x (30x − 10x² − 5x²) dx
dV = 2π x (30x − 15x²) dx
dV = 30π (2x² − x³) dx
The total volume is the sum (integral):
V = ∫ dV
V = ∫₀² 30π (2x² − x³) dx
V = 30π ∫₀² (2x² − x³) dx
V = 30π (⅔ x³ − ¼ x⁴)|₀²
V = 30π (⅔ 8 − ¼ 16)
V = 30π (16/3 − 4)
V = 10π (16 − 12)
V = 40π
Answer:
I dont know sorry i in a hurry and i need points so vary sorry
Explanation:
my assistant is due at 9:00 today . Not that you care but
Answer:
Complete question for your reference is below.
A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P = 19 kN are applied at the centers of the ends of the bar. Knowing that a = 28 mm and sigma_all = 130 MPa, determine the smallest allowable depth d of the milled portion of the bar
Explanation:
Please find attached file for complete answer solution and explanation.
