An automobile tire is rated to last for 60,000 miles. to an order of magnitude, through how many revolutions will it turn?

1 answer:

3 0

Let the average diameter of the automobile tire is about 2 feet. Therefore its circumference is $2 feet\times \pi =6.29 feet$ .

Now we have to convert an automobile tire is rated to last for 60,000 miles in to feet.

As $1 miles =5280 feet$ .

So, $60000 miles = 60000\times 5280 feet =316800000 feet$

Thus the number of revolutions,

$\frac{316800000 feet }{6.29 feet} =50365659.8 revolutions \\\\=5.0\times10^7 rev$

Hence, order of magnitude of revolutions is 7.

You might be interested in

To what temperature must you raise a silver wire (c = 0.0038), originally at 20.0°C, to double its resistance, neglecting any ch

balu736 [363]

Answer:

$T=283^{\circ}C$

Explanation:

Given a material with temperature coefficient of resistance <em>c</em>, the equation that relates the resistance $R_0$ at temperature $T_0$ and the resistance $R$ at temperature $T$ is

$\frac{R-R_0}{R_0}=c(T-T_0)$

We want to double our resistance, so $R=2R_0$ , thus having:

$\frac{2R_0-R_0}{R_0}=\frac{R_0}{R_0}=1=c(T-T_0)$

For this T must be:

$1=cT-cT_0$

$T=\frac{1+cT_0}{c}$

which for our values means (with $T=20^{\circ}C=293^{\circ}K$ , remember to write temperature in S.I., and that for silver $c=0.0038^{\circ}K^{-1}$ ):