Answer:
C
Explanation:
Definition of drug: a medicine or other substance which has a physiological effect when ingested or otherwise introduced into the body
A gold doubloon 6.1 cm in diameter and 2.0mm thick is dropped over the side of a Pirate Ship. When it comes to rest on the ocean floor at a depth of 770m how much has its volume changed
The volume of a material is the total amount of matter that it can contain. The volume of the given coin has been determined to be 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex]. Since the gold doubloon do not absorb water, then its volume remains constant at the ocean floor.
The volume of the gold doubloon can be determined by;
volume = [tex]\pi r^{2}[/tex] + h
where r is the radius of the coin and h is its thickness.
Such that; diameter = 6.1 cm (61 mm) and h = 2.0 mm
r = [tex]\frac{diameter}{2}[/tex]
= [tex]\frac{61}{2}[/tex]
r = 30.5 mm
Thus,
volume of the coin = [tex]\frac{22}{7}[/tex] x [tex](30.5)^{2}[/tex] x 2
= 5847.2857
Therefore, the volume of the gold doubloon is 5847.3 [tex]mm^{3}[/tex]. This can also be expressed as 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex].
Since the gold doubloon is not miscible with water, thus its volume at a depth of 770 m at the ocean floor is the same as its initial volume.
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b) A satellite is in a circular orbit around the Earth at an altitude of 1600 km above the Earth's surface. Determine the orbital period of the satellite in hours. [3]
Explanation:
The orbiting period of a satellite at a height h from earth' surface is
T=2πr32gR2
where r=R+h.
Then, T=2π(R+h)R(R+hg)−−−−−−−−√
Here, R=6400km,h=1600km=R/4
T=2πR+R4−−−−−−√R(R+R4g)−−−−−−−−−⎷=2π(1.25)32Rg−−√
Putting the given values,
T=2×3.14×(6.4×106m9.8ms−2)−−−−−−−−−−−−√(1.25)32=7092s=1.97h
Now, a satellite will appear stationary in the sky over a point on the earth's equator if its period of revolution around the earthh is equal to the period of revolution of the earth up around its own axis whichh is 24h. Let us find the height h of such a satellite above the earth's suface in terms of the earth,'s radius.
Let it be nR.Then
T=2π(R+nR)R(R+nRg)−−−−−−−−−−√
=2π(Rg)−−−−−√(1+n)32
=2×3.14(6.4×106m/s9.8m/s2)−−−−−−−−−−−−−−−⎷(1+n)32
(5075s)(1+n)32=(1.41h)(1+n)32
For T=24h, we have (24h)=(1.41h)(1+n)32
or (1+n)32=241.41=17
or 1+n(17)23=6.61
or n=5.61
The height of the geostationary satellite above the earth's surface is nR=5.61×6400km=6.59×104km.