Answer:
%Open the file.
fID = fopen('parts_inv.dat');
%Read from the file.
data = fscanf(fID,'%d\t%f\t%d',[3,inf]);
%Close
fclose(fID);
%Restore the data.
data = data';
%Get the size
[rs, cs] = size(data);
%Set value.
invCost = 0;
%Loop
for rw = 1 : rs
%Find cost
invCost = invCost + (data(rw, 2) * data(rw, 3));
%Loop end
end
%Display the cost.
fprintf('Total cost: %4.2f\n\n', invCost);
Explanation:
Answer:

Explanation:
From conservation of energy states that

Answer:
r₂ = 0.316 m
Explanation:
The sound level is expressed in decibels, therefore let's find the intensity for the new location
β = 10 log
let's write this expression for our case
β₁ = 10 log \frac{I_1}{I_o}
β₂ = 10 log \frac{I_2}{I_o}
β₂ -β₁ = 10 (
)
β₂ - β₁ = 10
log \frac{I_2}{I_1} =
= 3
= 10³
I₂ = 10³ I₁
having the relationship between the intensities, we can use the definition of intensity which is the power per unit area
I = P / A
P = I A
the area is of a sphere
A = 4π r²
the power of the sound does not change, so we can write it for the two points
P = I₁ A₁ = I₂ A₂
I₁ r₁² = I₂ r₂²
we substitute the ratio of intensities
I₁ r₁² = (10³ I₁ ) r₂²
r₁² = 10³ r₂²
r₂ = r₁ / √10³
we calculate
r₂ =
r₂ = 0.316 m
Explanation:
the object has constant velocity for 2 seconds and it get a constant accelration (2ms-2)