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earnstyle [38]
3 years ago
5

Oceanography in a sentence

Physics
2 answers:
pickupchik [31]3 years ago
7 0

Thus geology, meteorology, oceanography and anthropology developed into distinct sciences. Oceanography is the science which deals with the ocean, and since the ocean forms a large part of the earth's surface oceanography is a large department of geography.

GaryK [48]3 years ago
5 0

Oceanography is the branch of science that deals with the physical and biological properties and phenomena of the sea.

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Which property of matter is conserved in chemical reactions and shown by balanced equations?
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Answer:

Mass.

Explanation:

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Need helppp number 5
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1768 meters

Explanation:

Distance = speed x time.

(340 m/s) x (5.2 s) = 1768 m

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How much heat is needed to raise the temperature of 100 g of water by 50 C, if the specific heat of water is 4,184 J/kg.C?
Yuliya22 [10]

Heat required to raise the temperature of water is given as

Q = ms\Delta T

here we have

m = 100 g = 0.100 kg

s = 4183 J/kg C

\Delta T = 50 ^0 C

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Q = 0.100 * 4184 * 50

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so here it requires 20920 J heat to raise the temperature of 100 g water by 50 degree C

4 0
3 years ago
How does Newton's second law of motion can be used to calculate the acceleration of an object?
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4 years ago
You have a glass ball with a radius of 2.00 mm and a density of 2500 kg/m3. You hold the ball so it is fully submerged, just bel
Maurinko [17]

Answer:

(a) check attachment

(b)5 m/s²

Explanation:

Given: radius = 2.00mm: density = 2500kg/m³: viscosity of glycerin = 1.5pa: decity of glycerin = 1250kg/m³: g = 10N/kg = 10m/s²: Fdrag = 6πnrv

(a) for answer check attachment.

(b) For the magnitude of the balls initial acceleration:

     Initial net force(f) = mg - upthrust

                                  = mg - (\frac{m}{p} )pg.g

     acceleration (a) = Acceleration(a)=\frac{f}{m}\\=g - (\frac{pg}{p})g\\=g(1-\frac{pg}{p} )\\=10(1-\frac{1250}{2500} )\\a=10(1-0.5)\\a=5 m/s²

c.) fromthe force diagram in the attachment; when the ball attains terminal velocity the net force will be zero(0)

                                mg=6πnrv + upthrust

d.) For the magnitude of terminal velocity:

                                             mg=6πnrv + (\frac{m}{p})pg.g\\\\(\frac{4}{3}πr^{3} p)g=6πnrv +\frac{4}{3}πr^3pg.g\\\\V = \frac{2}{9}.\frac{(2*10^{-3})^{2}*(2500-1250)*10}{1.5}\\\\=0.79cm/s

e.) when the ball reaches terminal velocity, the acceleration is zero (0)

8 0
3 years ago
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