Investigative and mathematical skills in science EMA

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Investigative and mathematical skills in science EMA
Question 1
(a) We know that
y= 4πLxFor the first part of problem (i), we have to find the base units of y.
It is given that L has units of meters (m) and x has units of meters per second per second (m/s2). In order to find the value of y, the following expression can be simplified while neglecting the constants which is 4π.
We have
y= mms2=m2s2= msHence the unit of y is meters per second (m/s).
(ii) For the second part of the question, a straight line can only be obtained if the equation can be expressed in the form of
y = mx + c
where m is the slope of line and c is the y-intercept.
Rearranging the above equation, we have
y= 4πL x xHowever x not a linear is function and will surely give a curve. But, if the above equation is bounded for only positive squared values of x, the above function would be a straight line. However, the constant terms would be the slope of that straight line.
(iii) When we have a straight line, with the data of x and y. We have to find the slope which can be found by
Slope m= ∆y∆xConsidering the equation, the slope would be positive and once the slope is known, it can be compared with general line equation (shown above) to find the value of L. Mathematically;
m= 4πL Rearranging the equation will give,
L= m24πThe above equation will determine the value of constant, L.
(b) The graph is provided and tangent is drawn on it.

In order to find the rate of change of temperature with respect to time,
slope m= ∆Temperature∆Time= 0-82.

59.85-0(1000)= -0.008 ℃sec(ii) As given that,
T=a+be-ktDifferentiate above equation with respect to t, we have
dTdt=0+(be-kt)( -k)Simplifying above equation, we have
dTdt=-kbe-ktIt is given that b = 80oC and k = 2.4×10-4 sec-1.
For 2.0×103 seconds;
dTdtat t = 2000 sec=- 2.4×10-480e-2.4×10-42000= -0.011℃sec For 6.0×103 seconds;
dTdtat t = 6000 sec=- 2.4×10-480e-2.4×10-46000= -0.0045℃sec(iii) In the light of above functions, the graphs of each of the above curves are quite different from one another. The graph in part one, would have a curve with a positive slope. However, the exponential function would have a negative slope of curve with an increasing trend.
A
(c) The picture can be simplified as;
79o

b

Total Height (c) = 12.69 m

a
C
B

For a triangle, we know that
m∠A+m∠B+m∠C=180oWe know that ∠A = 79o and ∠B = 90o, so now we can find the value of ∠C; we get
m∠C=180o- 79o- 90o=11oApplying Sine law;
aSin (A)= bSin (B)= cSin (C)We have to find the distance “a” in this question and “b” is unknown, so above equation can be simplified as;
aSin (A)= cSin (C)Rearranging above equation in terms of “a”, we get;
a= cSin (C) x Sin A= 12.69Sin (11) x Sin 79= 65.3 mHence, the distance from surfer and the boy is 65.3 meters.
(d) For the sake of simplicity, consider the following diagram
b
Diameter of satellite (a) = 13 m
C

0.011o

B
c
A

In this diagram A is the observer, and for this problem, we have to find the distance “c”.
For a triangle, we know that
m∠A + m∠B + m∠C=180oWe know that ∠A = 0.011o and ∠B = 90o, so now we can find the value of ∠C; we get
m∠C=180o- 0.011o- 90o=89.98oApplying Sine law;
aSin (A)= bSin (B)= cSin (C)We have to find the distance “a” in this question and “b” is unknown, so above equation can be simplified as;
aSin (A)= cSin (C)Rearranging above equation in terms of “a”, we get;
a= cSin (C) x Sin A= 13Sin (0.011) x Sin 89.98= 67 kmHence, the distance from surfer and the boy is 67 kilometers.
(e) For the expression ln(5t4), the expressions “A” and “D” are equivalent as per the laws of logarithm.
The expression ln(5t4) can be further solved as
ln (5t4) = ln (5) + ln (t4) = ln (5) + 4ln (t)
(f) For calculation of mean and standard deviation following formulae (for ungrouped data) are used in excel spreadsheet
x= xn
σ= (x- x)2n