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Physics Bootcamp

Samuel J. Ling

Section 1.9 Introduction Bootcamp

Exercises Exercises

Units

4. Unit Conversion from MPH to Meters per Second.
Follow the link: Exercise 1.3.5.2.
5. Coverting mph to m/s.
A car travels at a speed of 75 miles per hour (mph). What is the speed in meters/second (m/s)?
Answer.
33.5 m/s.
7. Converting Light Years to Kilometers.
The star Betelgeuse is 650 light years from Earth. How far is it in kilometers?
Answer.
\(6.2 \times 10^{15}\text{ km}\text{.}\)
10. Conversion of Meters to Astronomical Unit Distance.
Astronomical unit (AU) is a measure of distance that is equal to the average distance between the Earth and the Sun. One AU is approximately equal to 150 million kilometers. Evaluate the average distances between the Sun and following planets, Mercury, Mars, and Jupiter, in AU if these distance in meter are \(5.79 \times 10^{10}\ \text{m}\text{,}\) \(2.28 \times 10^{11}\ \text{m}\) and \(7.78 \times 10^{11}\ \text{m}\) respectively.
Answer.
\(d_{S-Mer}=3.86\times10^{-1}\ \text{AU}\text{,}\) \(d_{S-Mar}=1.52\ \text{AU}\text{,}\) \(d_{S-J}=5.19\ \text{AU}\text{.}\)

Uncertainty, Significant Figures, Rounding

11. Illustration of Significant Figures.
Follow the link: Example 1.16.
15. Avoiding Speeding Ticket by Using Uncertainty.
Follow the link: Exercise 1.4.8.4.

Propagation of Uncertainty

16. Propagation of Uncertainties to Volume from Uncertainties in Length, Width, Height.
Follow the link: Example 1.18.
17. Absolute and Relative Uncertainties in Area of a Square.
Follow the link: Exercise 1.5.3.1.
18. Perimeter and Area of a Square from Measurements one Side.
The side of a square plate is measured to be \(5.0\ \text{cm}\pm 0.2\ \text{cm}\text{.}\) (a) What are the absolute and relative uncertainties in the measured value of the side? (b) Find the perimeter and the area of the square to the approprtiate significant figures. (c) Find the absolute and relative uncertainties in the perimeter and the area of the square.
Answer.
(a) \(0.2\ \text{cm}\text{,}\) \(0.04\) or \(4\%\text{.}\) (b) \(2.0\ \text{cm}\text{,}\) \(2.5\times 10^{1}\ \text{cm}^2\text{.}\) (c) Prerimeter: \(0.8\ \text{cm}\text{,}\) \(0.04\) or \(4\%\text{;}\) Area: \(2.0\ \text{cm}^2\text{,}\) \(0.08\) or \(8\%\text{.}\)
19. Absolute and Relative Uncertainties in Area of a Circle.
Follow the link: Exercise 1.5.3.2.
20. Measuring Diameter of a Circle for Uncertainty.
Use a ruler with mm markings to measure the diameter of the given circle in Figure 1.24, and calculate the average and the uncertainties of circumference and area. (Note: Your measurements will have an average value and an uncertainty.)
Figure 1.24.
22. Uncertainty in Mass from Measurements on Diameter.
Follow the link: Exercise 1.5.3.4.
23. Uncertainty in Volume of a Cylinder from Measurements on Diameter and Height.
Follow the link: Exercise 1.5.3.5.

Order of Magnitude

25. Estimating Number of Marbles in a Jar.
Follow the link: Example 1.19.
27. Estimating Amount of Blood Pumped in a Day.
Follow the link: Exercise 1.6.2.
28. Estimating Numer of Hair on Full Scapl.
Follow the link: Exercise 1.6.3.
29. Estimate the Total Mass of All Water in Earth’s Oceans.
Follow the link: Exercise 1.6.4.
30. Estimate Gasoline Used in the United States.
Follow the link: Exercise 1.6.5.

Dimensional Analysis

32. Checking Dimensions in a Physics Equation.
Follow the link: Example 1.23.
33. Guessing the Formula for Frequency of a Pendulum.
Follow the link: Exercise 1.7.2.2.
34. Predicting Formula for Priod of a Block Attached to a Spring.
Follow the link: Exercise 1.7.2.3.

Miscellaneous

35. Distance To Stars from Subtended Angle.
In Astronomy, a unit of distance, called parsec (pc), is in common use. It is defined to be the distance at which an object of size 1 Astronomical Units (AU) \((149.6\times10^6\ \text{km})\) will subtend an angle of 1 arc-second, which is equal to \(1/3600\) of one degree. Two stars in a binary star system separated by \(8.6 \times 10^{14}\ \text{m}\) are seen to subtend an angle of \(0.2\ \text{arcsec}\text{.}\) How far away are the stars (a) in pc, and (b) in meters.
Hint.
Answer.
(a) \(2.9 \times 10^4\text{ pc}\text{;}\) (b) \(8.9 \times 10^{20}\text{ m}\text{.}\)
Solution.
From the definition of parsec, we see that parsec is the value radius \(r\) in the formula
\begin{equation*} s = r \theta, \end{equation*}
when \(\theta = 1\,\text{arcsec}\) and \(s = 1\, \text{AU} = 149.6\times10^6\ \text{km}\text{.}\) Let us express \(\text{arcsec}\) in radians.
\begin{equation*} 1\,\text{arcsec} = \frac{1}{3600}\,\text{deg} = \frac{1}{3600} \times \frac{\pi}{180}\,\text{rad} = \frac{\pi}{648,000}\,\text{rad}. \end{equation*}
Therefore,
\begin{equation*} 1\,\text{pc} = \frac{s}{\theta} = \frac{149.6\times10^6\ \text{km}}{\frac{\pi}{648,000}\,\text{rad}} = 3.086\times 10^{13}\, \text{km}. \end{equation*}
Now, the distance between the two stars is the new \(s\) value for which \(\theta = 0.2\,\text{arcsec}\text{.}\) Hence, the distance \(r\) to the stars will be
\begin{equation*} r = \frac{s}{\theta} = \frac{8.6 \times 10^{14}\ \text{m}}{0.2\,\text{arcsec}}. \end{equation*}
Converting the arcsec to radian and carrying out the calculation gives us \(r\) in \(\text{km}\) to be
\begin{equation*} r = 8.87\times 10^{20}\,\text{m}. \end{equation*}
This cae=n be converted to
\begin{equation*} r = \frac{8.87\times 10^{20}\,\text{m}}{3.086\times 10^{16}\, \text{m/pc}} = 2.87\times 10^4\,\text{pc}. \end{equation*}
36. Estimate Radius of Earth from Dip of Sail.
A \(2\text{-m}\) pole is attached to a boat. When the boat is \(5\text{ km}\) away from the shore, you can see only tip of the pole. Use this information and assumption that Earth is a spherical ball to estimate the radius of earth.
Figure 1.25.
Hint.
Use Pythagora’s theorem.
Answer.
\(6.3\times 10^6\, \text{m}.\)
Solution.
From the geometry, you can identify a right-angled triangle with sides \(R\text{,}\) \(5000\,\text{m}\text{,}\) and \(R + 2\,\text{m}\text{.}\) Hence, we can set up a Pythogora’s equation and solve for \(R\text{.}\)
\begin{equation*} (R+2)^2 = R^2 + 5000^2. \end{equation*}
This gives
\begin{equation*} R = \frac{25,000,000 - 4}{4} = 6.3\times 10^6\, \text{m}. \end{equation*}
37. Estimate Radius of Exploding Ball.
The British physicist G. I. Taylor argued that the radius \(R\) of a spherically symmetric nuclear explosion must depend on the energy \(E\text{,}\) the initial density of air \(\rho\text{,}\) and time \(t\) since explosion. Using dimensional analysis, find a formula for the radius at time t after the explosion. [Challenging problem]
Hint.
Express all quantities into their fundamental units.
Answer.
\(R^5=K E \rho^{-1}t^{2}\text{,}\) where \(K\) is a dimensionless constant.
Solution.
Let us express the multiplicated relation as
\begin{equation*} R = K E^a \rho^b t^c, \end{equation*}
where \(K\) is a dimensionless constant, which cannot be determined by a dimensional analysis. Now, we express each side into their fundamental dimensions. We drop the constant from this part.
\begin{align*} \text{Left side} \amp = L^1 \\ \text{Right side} \amp = \left( \frac{M L^2}{T^2} \right)^a \left( \frac{M}{L^3} \right)^b T^c = M^{a + b} L^{2a-3b} T^{-2a + c} \end{align*}
Now, equating the exponents of same unit/dimension on both sides (noting the missing unit on one side refers to the exponent of that unit being zero) we get
\begin{align*} \amp a + b = 0 \\ \amp 2a - 3b = 1 \\ \amp -2a + c = 0 \end{align*}
These are readily solved to yield
\begin{equation*} a = 1/5,\ b = -1/5,\ c = 2/5. \end{equation*}
Hence,
\begin{equation*} R = K E^{1/5} \rho^{-1/5} t^{2/5}. \end{equation*}
This is suggestive that we should look \(>R^5\text{,}\) which will give
\begin{equation*} R^5 = K'\, \frac{E}{\rho}\,t^2, \end{equation*}
where \(K'\) is some domensionless constant.
38. Predict Formula for Oscillation Frequency of a Star.
By dimensional analysis, find a formula for the oscillation frequency of a star of radius \(R\) and density \(\rho\text{.}\) Note that you will also need the dimensions of Newton’s gravitational constant also, which is \([G_N]=[L]^3[T]^{-2}[M]^{-1}\text{.}\) [Challenging problem]
Hint.
Express all quantities into their fundamental units.
Answer.
\(= K \sqrt{G_N \rho}.\)
Solution.
We will assume a multiplicative relation such as
\begin{equation*} f = K G_N^a R^b \rho^c, \end{equation*}
where \(K\) is an undetermined dimensionless constant. Now, the representations of the two sides in fundamental units gives (we ignore \(K\) in this step.)
\begin{align*} T^{-1} \amp = \left( L^3T^{-2}M^{-1} \right)^a \times L^b \times \left( M L^{-3} \right)^c \\ a\mp = L^{3a + b -3c} M^{-a + c} T^{-2a} \end{align*}
Therefore, we have
\begin{equation*} -2a = -1,\ -a + c = 0,\ 3a + b - 3c = 0. \end{equation*}
These yield the following values.
\begin{equation*} a = c = 1/2, b = 0. \end{equation*}
Hence,
\begin{equation*} f = K \sqrt{G_N \rho}. \end{equation*}
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