Answer: 78.89%
Explanation:
Given : Sample size : n= 1200
Sample mean : 
Standard deviation : 
We assume that it follows Gaussian distribution (Normal distribution).
Let x be a random variable that represents the shaft diameter.
Using formula, 
, the z-value corresponds to 2.39 will be :-
z-value corresponds to 2.60 will be :-
Using the standard normal table for z, we have
P-value = 
Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%
Answer:
B. The thickness of the heated region near the plate is increasing.
Explanation:
First we know that, a boundary layer is the layer of fluid in the immediate vicinity of a bounding surface where the effects of viscosity are significant. The fluid is often slower due to the effects of viscosity. Advection i.e the transfer of heat by the flow of liquid becomes less since the flow is slower, thereby the local heat transfer coefficient decreases.
From law of conduction, we observe that heat transfer rate will decrease based on a smaller rate of temperature, the thickness therefore increases while the local heat transfer coefficient decreases with distance.
The C++ code that would draw all the iterations in the selection sort process on the array is given below:
<h3>C++ Code</h3>
#include <stdio.h>
#include <stdlib.h>
int main() {
int i, temp1, temp2;
int string2[16] = { 0, 4, 2, 5, 1, 5, 6, 2, 6, 89, 21, 32, 31, 5, 32, 12 };
_Bool check = 1;
while (check) {
temp1 = string2[i];
temp2 = string2[i + 1];
if (temp1 < temp2) {
string2[i + 1] = temp1;
string2[i] = temp2;
i = 0;
} else {
i++;
if (i = 15) {
check = !check;
}
}
}
return 0;
}
Read more about C++ programming here:
brainly.com/question/20339175
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Answer:
E= 15 GPa.
Explanation:
Given that
Length ,L = 0.5 m
Tensile stress ,σ = 10.2 MPa
Elongation ,ΔL = 0.34 mm
lets take young modulus = E
We know that strain ε given as
We know that
Therefore the young's modulus will be 15 GPa.
Answer:
0.023 Pa*s
Explanation:
The surface area of the side of the inner cylinder is:
A = π*d*l
A = π*0.15*0.75 = 0.35 m^2
At 200 rpm the inner cylinder has a tangential speed of:
u = w * r
u = w * d/2
w = 200 rpm * 2π / 60 = 20.9 rad/s
u = 20.9 * 0.15 / 2 = 1.57 m/s
The torque is of 0.8 N*m, this means that the force is:
T = F * r
F = T / r
F = 2*T / d
For Newtoninan fluids with two plates moving respect of each other with a fluid between the viscous friction force would be:
F = μ*A*u / y
Where
μ: viscocity
y: separation between pates
A: surface area of the plates
Then:
2*T / d = μ*A*u/y
Rearranging:
μ = 2*T*y / (d*A*u)
μ = 2*0.8*0.0012 / (0.15*0.35*1.57) = 0.023 Pa*s
