Convert the following to SI units:
a) 8.0 in
b) 66 ft/s
c) 60 mph
d) 14 in2
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PROBLEM:
SOLUTION:
We assume that the earth revolves around the sun in a circular manner. Therefore, the distance between the earth and the sun will be the radius of its orbit.
The total distance traveled the earth in full revolution is
\begin{align*} d & =2\pi r \\ & = 2\pi \left(10\right)^8\:\text{km} \\ \end{align*}
The total time of travel is
\begin{align*} t & =365.25\:\text{days}\left(\frac{24\:\text{hr}}{1\:\text{day}}\right)\left(\frac{3600\:\text{s}}{1\:\text{hr}}\right) \\ & = 3.15576\times 10^7\:\text{sec} \\ \end{align*}
Therefore, the average speed is
\begin{align*} \text{speed} & =\frac{\text{distance}}{\text{time}}=\frac{d}{t} \\ & = \frac{2\pi \left(10\right)^8\:\text{km}}{3.15576\times 10^7\:\text{s}} \\ & = 19.91\:\text{km/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
Therefore, the average speed of the earth is 19.91 km/s.
Convert 19.91 km/s to m/s.
\begin{align*} s & =\left(19.91\:\frac{\text{km}}{\text{s}}\right)\left(\frac{1000\:\text{m}}{1\:\text{km}}\right) \\ & =19\,910\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}
Therefore, the velocity is 19 910 m/s.
PROBLEM:
SOLUTION:
We know that the formula for the distance traveled, d, if we are given rate, r, and time, t is given by d=r\times t.
The rate in cm/s is
\begin{align*} r & =\left(4\:\frac{\text{cm}}{\bcancel{\text{year}}}\times \frac{1\:\bcancel{\text{year}}}{365\:\frac{1}{4}\:\bcancel{\text{days}}}\times \frac{1\:\bcancel{\text{day}}}{24\:\bcancel{\text{hours}}}\times \frac{1\:\bcancel{\text{hour}}}{60\:\bcancel{\text{mins}}}\times \frac{1\:\bcancel{\text{min}}}{60\:\text{s}}\right) \\ & = 1.2675 \times 10^{-7} \ \text{cm/s} \end{align*}
The distance traveled is
\begin{align*} d & = r \times t \\ d & = \left( 1.2675\times 10^{-7} \ \text{cm/s} \right)\left( 1.0 \ \text{s} \right) \\ d & = 1.3\times 10^{-7} \ \text{cm} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
We need to convert 4.0 cm/year to km/My.
\begin{align*} 4.0 \ \frac{\text{cm}}{\text{year}} & = 4.0\:\frac{\bcancel{\text{cm}}}{\bcancel{\text{year}}}\times \frac{1\:\bcancel{\text{m}}}{100\:\bcancel{\text{cm}}}\times \frac{1\:\text{km}}{1000\:\bcancel{\text{m}}}\times \frac{1\:000\:000\:\bcancel{\text{years}}}{1\:\text{My}} \\ \\ & =40\:\text{km/My} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
Therefore, the speed of the tectonic plates is 40 km/My.
PROBLEM:
SOLUTION:
We know that 1 km=1000 m and 1 hr=3600 sec.
Convert 342 m/s.
\begin{align*} 342\:\text{m/s} & =\left(342\:\frac{\bcancel{\text{m}}}{\bcancel{\text{s}}}\right)\left(\frac{1\:\text{km}}{1000\:\bcancel{\text{m}}}\right)\left(\frac{3600\:\bcancel{\text{s}}}{1\:\text{hr}}\right) \\ & =1231.2\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
Therefore, the speed of sound in km/hr is 1231.2 km/hr.
PROBLEM:
SOLUTION:
Convert 29028 ft to km
29,028\:\text{ft}=\left(29\:028\:\text{ft}\right)\left(\frac{1\:\text{km}}{3281\:\text{ft}}\right)=8.847\:\text{km} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
Therefore, the height of Mount Everest is 8.847 kilometers.
PROBLEM:
SOLUTION:
First, convert 6 ft to inches
6\:\text{ft}=\left(6\:\text{ft}\right)\left(\frac{12\:\text{in}}{1\:\text{ft}}\right)=72\:\text{in}
Add the 1.0 inch
72\:\text{in}\:+1\:\text{in}=73\:\text{inches}
So, the total height of the person is 73 inches. We convert this to meters to come up with the desired unit.
73\:\text{in}=\left(73\:\text{in}\right)\left(\frac{1\:\text{m}}{39.37\:\text{in}}\right)=1.85\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
So, the height of the person is 1.85 meters.
PROBLEM:
SOLUTION:
The length in feet and inches are
\begin{aligned} 115\ \text{m} & = 115\ \bcancel{\text{m}} \times \frac{1\ \text{ft}}{0.3048\ \bcancel{\text{m}}} \\ \\ & =377.3\ \text{feet} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ 115\ \text{m} & = 115\ \bcancel{\text{m}} \times \frac{1\ \bcancel{\text{ft}}}{0.3048\ \bcancel{\text{m}}} \times \frac{12\ \text{inches}}{1\ \bcancel{\text{ft}}} \\ \\ & =4528\ \text{inches} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{aligned}
The width in feet and inches are
\begin{aligned} 85\ \text{m} & = 85\ \bcancel{\text{m}} \times \frac{1\ \text{ft}}{0.3048\ \bcancel{\text{m}}}\\ \\ & =278.9\ \text{ft}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ 85\ \text{m} & = 85\ \bcancel{\text{m}} \times \frac{1\ \bcancel{\text{ft}}}{0.3048\ \bcancel{\text{m}}} \times \frac{12\ \text{in}}{1\ \bcancel{\text{ft}}}\\ \\ & =3346\ \text{inches} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{aligned}
PROBLEM:
SOLUTION:
\begin{aligned} 100 \ \text{yard} & = 100 \ \bcancel{\text{yard}} \times \frac{3\ \bcancel{\text{feet}}}{1 \ \bcancel{\text{yard}}}\times \frac{1 \ \text{m}}{3.281\ \bcancel{\text{feet}}} \\ \\ & =91.4 \ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{aligned}
PROBLEM:
SOLUTION:
We know that 1 hr = 3600 s, and 1 km = 1000 m.
\begin{aligned} 1.0 \ \text{m/s} & = 1.0 \ \frac{\bcancel{\text{m}}}{ \bcancel{\text{s}}} \times \frac{3600 \ \bcancel{\text{s}}}{1 \ \text{hr}}\times \frac{1 \ \text{km}}{1000 \ \bcancel{\text{m}}} \\ \\ & =3.6 \ \text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{aligned}
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