How do you increase capacitance in a capacitor?
1 answer:
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Maximum error in the result of the sum of measurement is equal to the sum absolute error of the individual observed measurements
(a) The maximum error in D is 0.09
(b) The standard error in D is approximately 0.034
The procedure for arriving at the above values is as follows;
The given measurements and the sited errors are;
A = 15.21 + 0.01
B = 10.82 + 0.05
C = 11.00 + 0.03
D = A + B + C
(a) Required parameter;
To calculate the maximum error in D
The equation for the propagation of error in addition is presented as follows;
Given that we have;
x = a + b
Therefore;
x + ±Δx = (a ± Δa) + (b ± Δb) = a + b ± (Δa + Δb)
∴ Δx = Δa + Δb
From the above formula, we have;
Where;
D = A + B + C
The maximum error in D = The sum of the maximum error in A, B, C
∴ The maximum error in D = 0.01 + 0.05 + 0.03 = 0.09
(b) Required parameter:
To find the standard error in D
The standard error is the sampling distribution's standard deviation, SD
Variance = SD²
The combined variance, SD² = The sum of the squares of individual standard deviations
Given that the standard errors represents the standard deviation, we get;
The combined variance, SD² = 0.01² + 0.05² + 0.03²
The combined variance, SD = √(0.01² + 0.05² + 0.03²) = 0.059
Where n = 3, for the three measurement, we get;
The standard error in D is approximately 0.034
Learn more about maximum error and standard error here:
Answer:
95.3 N
Explanation:
The tension in the cable is found by the equation:
Where is the mass density, and
is velocity.
First we find mass density:
-->
is mass:
and
is length of the cable:
, so:
And the velocity:
the time is and in that time the pulse went down and back along the cable 4 times, if one time down and back is:
2*4.15m=8.3m,
four times this path is:
4*8.3m=33.2m
thus, the velocity is:
And with this data we can now calculate the tension:
The tension is 95.3N