The first one is d or the 4th answer choice and the second one is false. Hope this helps!
Answer:
Please explain it in English so that i can help or you need someone else who can speak Vietnam
Answer:
diameter of the sprue at the bottom is 1.603 cm
Explanation:
Given data;
Flow rate, Q = 400 cm³/s
cross section of sprue: Round
Diameter of sprue at the top
= 3.4 cm
Height of sprue, h = 20 cm = 0.2 m
acceleration due to gravity g = 9.81 m/s²
Calculate the velocity at the sprue base
= √2gh
we substitute
= √(2 × 9.81 m/s² × 0.2 m )
= 1.98091 m/s
= 198.091 cm/s
diameter of the sprue at the bottom will be;
Q = AV = (π
/4) × 
= √(4Q/π
)
we substitute our values into the equation;
= √(4(400 cm³/s) / (π×198.091 cm/s))
= 1.603 cm
Therefore, diameter of the sprue at the bottom is 1.603 cm
Answer:
Code in MATLab is given as below:
Explanation:
grade = input('Enter the grades as elements of a vector ');
x1 = length(grade);
fprintf('There are %5.2f grades\n',x1);
x2 = mean(grade);
fprintf('The average grade is %5.2f \n',x2);
x3=std(grade);
fprintf('The standard deviation is %5.2f \n',x3);
x4 = median(grade);
fprintf('The median grade is %5.2f \n',x4);