Suppose that, while lying on a beach near the equator watching the Sun set over

Question

Suppose that, while lying on a beach near the equator watching the Sun set over a calm ocean, you start a stopwatch just as the top of the Sun disappears. You then stand, elevating your eyes by a height H = 1.75 m, and stop the watch when the top of the Sun again disappears. If the elapsed time is t = 12.9 s, what is the radius r of Earth? m The Sun disappears when it lies along a tangent to the Earth's surface and through your eyes. Draw a tangent for each position of your eyes. For the second sunset, draw a right triangle involving Earth's radius and your eye height in the elevated position. The triangle involves the angle through which Earth rotates in the given time interval. Grains of fine California beach sand are approximately spheres with an average radius of 50 pm and are made of silicon dioxide. A solid cube of silicon dioxide with a volume of 1.00 m3 has a mass of 2600 kg. What mass of sand grains would have a total surface area (the total area of all the individual spheres) equal to the surface area of a cube 1.20 m on an edge? Density is the mass per unit volume. Assume that the density of silicon dioxide is that of a sand grain. A sphere has surface area 4pir2 and volume (4/3)pir3. Cobalt has a density of 8.9 g/cm3, and the mass of a cobalt atom is 9.79 Times 10-26 kg. If the atoms are spherical and tightly packed, what is the volume of a cobalt atom? m3 What is the distance between the centers of adjacent atoms? m Gold, which has a mass of 19.32 g for each cubic centimeter of volume, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber. If a sample of gold, with a mass of 60.79 g, is pressed into a leaf of 2.200 pm thickness, what is the area of the leaf? If, instead, the gold is drawn out into a cylindrical fiber of radius 1.000 pm, what is the length of the fiber? First find the volume of the sample. Assuming the volume is constant, find the area in the flat shape and the length in the cylindrical state

Explanation / Answer

see in 11) a. we have 19.32 gm wet for 1cubic cm. so for 60.79 we would have (60.79)/(19.32) cm^3.

now as we know voume = area *depth.

so simply dividing the depth we can have the area.

so area A = (60.79)/(19.32*2.2*10^-4) cm^2

= 1.43 m^2

b. cylindrical fibre has volume pi*r^2*L where r is the radius and L is the length.

as we derived above volume = (60.79)/(19.32) cm^3.

so L = (60.79)/(19.32)/(pi*r^2) cm

L= 1001556 m

13. this simple as for 1 cubic metre it is 2600kg then for a cube edge 1.2m or the the voume 1.2^3 would have 2600*(1.2)^3 kg mass.

or = 4492.8kg.

14)

we know density = (mass/volume)

so here V = ((9.79*10^-23)/8.96) cm^3

= 1.1 *10^-29 m^3

b.

assuming the the atoms to be spherical

then we have 4/3 *pi*r^3 = V where r is the radius.

the distance of the centres is 2r. so equating both we get

distance = 2.759*10^-10 m

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.