please answer all part clearly and in details showing steps and equation numeric
ID: 2077496 • Letter: P
Question
please answer all part clearly and in details showing steps and equation
numerical solutions are given for B AND C SHOW HOW THEY WERE OBTAINED
(a) Discuss briefly how uncertainty propagation is analysed when multiple independent measurements with specified uncertainties are used to calculate a single measured quantity and its associated uncertainty.
(answer this)
(b) The coefficient of restitution (CR) of a ball can be determined by dropping the ball from a known height, H, onto a surface and then measuring its return height, h. For a particular experiment, CR is defined as CR = (h/H)0.5. If the uncertainty in height measurement is ± 1 mm and H = 1.0 m and h = 0.75 m, determine the coefficient of restitution and its associated uncertainty.
NUMERICAL ANSWER ((b) 0.866 ± 0721x10-3 )
(c) An electric kettle has a nominal power rating of 2875 W, based on a RMS supply voltage of 230 V and a current of 12.5 A. The power drawn by the kettle can be determined using the relation P = VI with reference to a voltmeter and an ammeter. Stating any assumptions, calculate the maximum tolerable uncertainties associated with the voltmeter and ammeter measurements, if the overall power consumption is to be measured to within an overall uncertainty of ±1%.
NUMERICAL ANSWER ((c) To achieve an overall measurement uncertainty that does not exceed ±1% of the nominal 2875 W value, the voltmeter and ammeter must have a precision uncertainties that do not exceed ±1.63 V and ±0.088 A, respectively.)
Explanation / Answer
a) When Y = f(a,b,c,...), and a=a0±a, b=b0±b, c=c0±c, ..., then the error propagation is defined as
Y2 = [(f/a)2 a2 + (f/b)2 b2 + (f/c)2 c2 + ...] evaluated at (a=a0, b=b0, c=c0, ...)
b) When CR = (h/H)^0.5 = (0.75/1)^0.5 = 0.866
CR2 = [(CR/h)2 h2 + (CR/H)2 H2 ] = H0.5 (0.5) h-0.5 h2 + h0.5 (-0.5) H-1.5h2
=> CR2 = (10.5 (0.5) 0.75-0.5)2 0.012 + (0.750.5 (-0.5) 1-1.5 )20.012 = 0.721 x10-3
So, CR = 0.866 ± 0.721 ×10-3
c) P = VI => P2 = (P/V)2 V2 + (P/I)2 I2= I2 V2 + V2I2
Uncertainly in power P = 1% of P = 0.01 × 2875 = 28.75
=> 28.752 = 12.52 V2 + 2302I2
Assuming that the uncentainly due to the voltage and current component will be same. Then
12.52 V2 = 28.752/2 => V = 1.63
Similarly, 2302I2 = 28.752/2 => I = 0.088
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