13) A space shuttle orbits the earth at an altitude of 300 km. What is the dista

Question

13) A space shuttle orbits the earth at an altitude of 300 km. What is the distance in (a) miles and (b)in millimeters? 14) The earth is approximately a sphere of a radius 6.37 x 10 6 m. What is the circumference of the earth in kilometers? What is the surface area in square kilometers? What is the earth's volume in cubic ft. 15) Antarctica is a roughly semicircle in shape with a radius of 2000 km. The average thickness of the ice cover is 3000 m. How many cubic centimeters of ice does Antarctica contain? (You can ignore the curvature of the earth) 16) Raindrops fall to earth from a cloud 1700 m above the earth's surface. Each water drop is approximately 20 milligrams. Assume that the drop is not slowed by air resistance, how fast would the drops be moving when they strike the ground? What force (in Newtons) will the drop have when it impacts the ground? Would it be safe to walk outside during a rainstorm under these conditions? 591 words

Explanation / Answer

Q 13 (a)Altitude of shuttle h = 300 km = 300 × 0.621 miles = 186.3 miles ( 1 km = 0.621 miles)

(b) h = 300 × 1000 × 1000 = 3 × 108 mm

Q.14. Radius of earth r = 6.37 × 106m

Circumference of earth C = 2× 22× r/ 7 = 40.0036 × 10 3 km

Surface area of earth s= 4 × 22 × r2 / 7 = 509.64 × 10 6 km2

volume of earth = v = 4× 22× r3/ (3 ×7 )= 1082.15 × 10 9 km3 = 1082.15 × 109 × 3.531 × 1010 cubic feet = 3.821 × 1019 cubic feet

Q.15. Radius of semicircle R = 2000 × 103m, thickness t = 3000 m

Volume of ice

V = area of semicircle × thickness = (22/7) R2 t /2

V= 18.84 × 109 CUBIC METER = 18.84 × 1015 cubic centimter

Q16.

Let u = initial velocity of drop = 0

Let v = final velocity

g = acceleration due to gravity = 9.8 ms-2

h = height through which drop falls = 1700 m

Using v2 = u2 + 2gh

v2= 66640

v = 258.15 ms-1

Force applied of rain drop assuming that final drop comes to rest in t = 0.01 s, = f = change in momentum of drop / time taken for this change = m ( 0 - v)/t = 0.2 × 10 -6 × 258.15/ 0.01 = 51.6 × 10-4 N

This is very small force. But pressure exerted by it would be large to our body hence it would be unsafe to walk through storm.

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