This question is incomplete, the complete question is;
A large tower is to be supported by a series of steel wires. It is estimated that the load on each wire will be 11,100 N.
Determine the minimum required wire radius assuming a factor of safety of 3 and a yield strength of 1500 MPa.
answer in mm please
Answer:
the minimum required wire radius is 5.3166 mm
Explanation:
Given that;
Load F = 11100N
N = 3
∝y = 1500 MPa
∝workmg = ∝y / N = 1500 / 3 = 500 MPa
now stress of Wire:
∝w = F/A
500 × 10⁶ = 11100 / A
A = 22.2 × 10⁻⁶ m²
so
(π/4)d² = A
(π/4)d² = 22.2 × 10⁻⁶
d² = 2.8265 × 10⁻⁵
d = 5.3165 7 × 10⁻³ m³
now we convert to mm(millimeters)
d = 5.3166 mm
Therefore the minimum required wire radius is 5.3166 mm
Answer:
The membrane diffusivity would be 3.968 x /s
Explanation:
According to Fick's law of diffusion, the rate of the purified product is related with to membrane diffusivity with the expression in equation 1.
..............................1
Where
V is the rate of purified product = 2 kg/hr = 2 kg/hr x 1 hr / 3600 sec = 1/1800 kg/sec
is the pressure at the supply gas point = 0.75 kg/
is the pressure at the take-off side = 0.05 kg/
A is the area of the membrane = 100
T is the thickness of the membrane = 0.05 mm = 0.05/ 1000 = 5 x m
Substituting the values into equation 1 we have;
5 x = 126000 x D
D = 5 x / 126000
D = 3.968 x /s
Therefore the membrane diffusivity would be 3.968 x /s
Answer:
Explanation:
Let assume that heating process occurs at constant pressure, the phenomenon is modelled by the use of the First Law of Thermodynamics:
The specific enthalpies are:
Liquid-Vapor Mixture:
Saturated Vapor:
The thermal energy per unit mass required to heat the steam is:
(a) If a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.
(b) For a kitten to be at 90th percentile, the minimum weight is 146.45 g.
<h3>
Weight distribution of the kitten</h3>
In a normal distribution curve;
- 2 standard deviation (2d) below the mean (M), (M - 2d) is at 2%
- 1 standard deviation (d) below the mean (M), (M - d) is at 16 %
- 1 standard deviation (d) above the mean (M), (M + d) is at 84%
- 2 standard deviation (2d) above the mean (M), (M + 2d) is at 98%
M - 2d = 125 g - 2(15g) = 95 g
M - d = 125 g - 15 g = 110 g
95 g is at 2% and 110 g is at 16%
(16% - 2%) = 14%
(110 - 95) = 15 g
14% / 15g = 0.93%/g
From 95 g to 99 g:
99 g - 95 g = 4 g
4g x 0.93%/g = 3.72%
99 g will be at:
(2% + 3.72%) = 5.72%
Thus, if a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.
<h3>Weight of the kitten in the 90th percentile</h3>
M + d = 125 + 15 = 140 g (at 84%)
M + 2d = 125 + 2(15) = 155 g ( at 98%)
155 g - 140 g = 15 g
14% / 15g = 0.93%/g
84% + x(0.93%/g) = 90%
84 + 0.93x = 90
0.93x = 6
x = 6.45 g
weight of a kitten in 90th percentile = 140 g + 6.45 g = 146.45 g
Thus, for a kitten to be at 90th percentile, the approximate weight is 146.45 g
Learn more about standard deviation here: brainly.com/question/475676
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Answer:
Dual-Mode Hybrid
Explanation:
The type of hybrid vehicle propelled by only an electric motor and does not require a traditional transmission to drive the wheels is known as "DUAL-MODE HYBRID."
Dual-Mode Hybrid is a type of Hybrid Electric Vehicle (HEV) which contains a separate generator consisting of rechargeable batteries. The engine ensures the wheels and the generator are moved; thereby, the electric motor and the batteries are fully powered.
A good example is a Toyota Prius, where during driving conditions, only the electric motor drives the wheels, in which the batteries supply the car with power.