Answer:
σ =5.39Mpa
Explanation:
step one:
The flexure strength is defined as the tendency with which unreinforced concrete yield to bending forces
Flexural strength test Flexural strength is calculated using the equation:
σ = FL/ (bd^2 )----------1
Where
σ = Flexural strength of concrete in Mpa
F= Failure load (in N).
L= Effective span of the beam
b= Breadth of the beam
step two:
Given data
F=40.45 kN= 40450N
b=0.15m
d=0.15m
L=0.45m
step three:
substituting into the expression we have
σ = 40450*0.45/ (0.15*0.15^2 )
σ =18202.5/ (0.15*0.15^2 )
σ =18202.5/ (0.15*0.0225 )
σ =18202.5/0.003375
σ =5393333.3
σ =5393333.3/1000000
σ =5.39Mpa
Therefore the flexure strength of the concrete is 5.39Mpa
Answer:
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Explanation:
Answer:
Yes,If we use river water which is entering at 20⁰ C in the condenser then it is possible to maintain the pressure of 10 KPa in condenser.
Explanation:
Yes,If we use river water which is entering at 20⁰ C in the condenser then it is possible to maintain the pressure of 10 KPa in condenser.
The saturation temperature of steam is 45.81⁰ C at the pressure of 10 KPa which is higher than 20⁰C of river water. So river water at 20⁰C can be used to maintain the condenser pressure to 10 KPa.
(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:
search up factors of 2450 and divide by two
