Using q=1.60217662 E-19 C we get:
Then using 9.10938356E-31 kg as the mass, we get
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:
The answer is Top-Down processing
Explanation:
I had this question on a apex quiz and i got it correct.
Answer:
271.095 m
Explanation:
✓ Let speed of sound in air that was given as (343 m/s) be represented as (Vi)
✓( speed of sound in concrete that was given as (3000 m/s ) be debited as (Vc)
✓ Let the distance travelled by the sound = s
✓duration of Time that exist between heard of sounds = 0.70s
But we know that
Time = (Distance / Speed)
✓Time it takes the sound to travel through air= s/vi = s/343
✓Time it takes the sound to travel through concrete= s/vc = s/3000
✓ (s/343) - (s/3000) = 0.70
Finding LCM and simplify
[(3000s - 343s)]/1029000 = 0.70
2657s /1029000 = 0.70
Making " s" subject of the formula
s= (1029000 × 0.70)/2657
s=720300/ 2657
s= 271.095 m
Hence, The impact took place at a distance of 271.095 m away from the person.
