Question

Q. 1. Earth is approximately a sphere of radius 6.37× 106 m.
What are (a) its circumference in kilometres, (b) its surface area in
square kilometres, and (c) its volume in cubic kilometres?
Q.2 The fastest growing plant on record is a Hesperoyucca whipplei that grew 3.7 m in 14 days. What was its growth rate in micrometres per second?

Answer 1

Answer:

Various geometric formulas are given in Appendix E.

Expressing the radius of the Earth of the Earth as R=(6.37×10

6

m)(10

−3

km)=6.37×10

3

km,

Its circumference is s=2πR=2π(6.37×10

3

km)=4.00×10

4

km.

Answer 2

Given: $r=6.37\times10^{6}m.$

We have to find the circumference, surface area, and volume of the earth.

1. (a)

$\text { The radius of the Earth of the Earth as } \mathrm{R}\\=>(6.37 \times\left.10^{6} \mathrm{~m}\right)\left(10^{-3} \mathrm{~km}\right)=6.37 \times 10^{3} \mathrm{~km}\\\text { Its circumference is } \mathrm{s}=>2 \pi \mathrm{R}=2 \pi\left(6.37 \times 10^{3} \mathrm{~km}\right)\\=>4.00 \times 10^{4} \mathrm{~km} \text {. }$

$\text { (b) surface area }=S A=4 \pi r^{2}\\=>4(3.14)(6.37\times10^{3}) ^{2} \\=>5.10\times10^{8} km^{2}$

(c) Volume$=\frac{4}{3} \pi.r^{3}$

$=>\frac{4}{3} (3.14)(6.37\times10^{3}) ^{3} \\\\=>1.082\times10^{12}km^{3}$

2)

$\text { We have to recognize the ratio, and then convert days into seconds. }\\\\\begin{array}{l}=>\left(\frac{14 \text { days }}{1}\right) \cdot\left(\frac{24 \text { hours }}{1 \text { day }}\right) \cdot\left(\frac{60 \text { minutes }}{1 \text { hour }}\right) \cdot\left(\frac{60 \text { seconds }}{1 \text { minute }}\right) \\\\=1209600 \text { seconds }\end{array}$

$\text { Then to convert meters into } \mu \mathrm{m}\\\left(\frac{3.7 \mathrm{~m}}{1}\right) \cdot\left(\frac{10^{6} \mu \mathrm{m}}{1 \mathrm{~m}}\right)=3700000 \mu \mathrm{m}\\\\\frac{3700000 \mu \mathrm{m}}{1209600 \mathrm{~s}} \approx 3.1 \frac{\mu \mathrm{m}}{\mathrm{s}}\\\\\text { The Hesperoyucca Whipplei growth rate during this period was } \approx 3.1 \frac{\mu \mathrm{m}}{\mathrm{s}}$