Answer:
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Explanation:
Answer:
![\dot m = 2.74 kg/s](https://tex.z-dn.net/?f=%5Cdot%20m%20%3D%202.74%20kg%2Fs)
Explanation:
given data:
pressure 1 MPa
diameter of pipe = 30 cm
average velocity = 10 m/s
area of pipe![= \frac[\pi}{4}d^2](https://tex.z-dn.net/?f=%20%3D%20%5Cfrac%5B%5Cpi%7D%7B4%7Dd%5E2)
![= \frac{\pi}{4} 0.3^2](https://tex.z-dn.net/?f=%20%3D%20%5Cfrac%7B%5Cpi%7D%7B4%7D%200.3%5E2)
A = 0.070 m2
WE KNOW THAT mass flow rate is given as
![\dot m = \rho A v](https://tex.z-dn.net/?f=%5Cdot%20m%20%3D%20%5Crho%20A%20v)
for pressure 1 MPa, the density of steam is = 4.068 kg/m3
therefore we have
![\dot m = 4.068 * 0.070* 10](https://tex.z-dn.net/?f=%5Cdot%20m%20%3D%204.068%20%2A%200.070%2A%2010)
![\dot m = 2.74 kg/s](https://tex.z-dn.net/?f=%5Cdot%20m%20%3D%202.74%20kg%2Fs)
Answer:per minute from the pumping well, a steady state was attained in about 24 hr. The draw-down at a distance of 10 ft. was 5.5 ft. and at 25 ft. was 1.21 ft.
Explanation:
Answer:
F = 0.0022N
Explanation:
Given:
Surface area (A) = 4,000mm² = 0.004m²
Viscosity = µ = 0.55 N.s/m²
u = (5y-0.5y²) mm/s
Assume y = 4
Computation:
F/A = µ(du/dy)
F = µA(du/dy)
F = µA[(d/dy)(5y-0.5y²)]
F = (0.55)(0.004)[(5-1(4))]
F = 0.0022N
Answer:
1.887 m
Explanation:
(15 *pi)/180
= 0.2618 rad
Polar moment
= Pi*d⁴/32
= (22/7*20⁴)/32
= 15707.96
Torque on shaft
= ((22/7)*20³*110)/16
= 172857.14
= 172.8nm
Shear modulus
G = 79.3
L = Gjθ/T
= 79.3x10⁹x(1.571*10^-8)x0.2618/172.8
= 1.887 m
The length of the bar is therefore 1.887 meters