If the volume of the gas increased from 600 to 800 in.^3, by how much does the pressure change?
2 answers:
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Answer:
a)
And since the power can't be negative then the solution would be w = 33.34 watts.
b)
So then the range of voltage would be between 33.22 W and 33.45 W.
c)For this case since that's the tolerance 1C
d)
So then we select the smalles value on this case
e) For this case if we assume a tolerance of for the temperature and a tolerance for the power input
we see that:
Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C
Step-by-step explanation:
For this case we have the following function
Where T represent the temperature in Celsius and w the power input in watts.
Part a
For this case we need to find the value of w that makes the temperature 203C, so we can set the following equation:
And we can rewrite the expression like this:
And we can solve this using the quadratic formula given by:
Where a =0.1, b =2.156 and c=-183. If we replace we got:
And since the power can't be negative then the solution would be w = 33.34 watts.
Part b
For this case we can find the values of w for the temperatures 203-1= 202C and 203+1 = 204 C. And we got this:
So then the range of voltage would be between 33.22 W and 33.45 W.
Part c
For this case since that's the tolerance 1C
Part d
For this case we can do this:
So then we select the smallest value on this case
Part e
For this case if we assume a tolerance of for the temperature and a tolerance for the power input
we see that:
Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C