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Reika [66]
4 years ago
15

Is the Pacific Ocean growing larger?

Physics
2 answers:
Ksivusya [100]4 years ago
6 0
Oceans<span> cover 70% of the Earth's surface. ... The Atlantic </span>ocean<span> and the Arctic </span>ocean<span>are good examples of active, </span>growing<span> oceanic </span>basins<span>, whereas the Mediterranean Sea is </span>shrinking<span>.</span>
Natalka [10]4 years ago
4 0
Oceans<span> cover 70% of the Earth's surface. The Atlantic </span>ocean<span> and the Arctic </span>ocean <span>are good examples of active, </span>growing<span> oceanic </span>basins<span>, whereas the Mediterranean Sea is </span>shrinking<span>. The </span>Pacific Ocean<span> is also an active, </span>shrinking<span> oceanic </span>basin<span>, even though it has both spreading ridge and oceanic trenches.</span>
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No links please, links never work, this is science not physics
stiks02 [169]

Answer:

b and c are the answers

A is an opinion, D is a superstitious belief because they haven’t found Jesus obviously

7 0
3 years ago
a swimmer can swim in still water at a speed of 9.50 m/s. he intends to swim directly across the river that has a downstream cur
Doss [256]
Refer to the diagram shown below.

Still-water speed  = 9.5 m/s
River speed = 3.75 m/s down stream.

The velocity of the swimmer relative to the bank is the vector sum of his still-water speed and the speed of the river.

The velocity relative to the bank is
V = √(9.5² + 3.75²) = 10.21 m/s

The downstream angle is
θ = tan⁻¹ 3.75/9.5 = 21.5°

Answer:  10.2 m/s at 21.5° downstream.

7 0
4 years ago
Read 2 more answers
PLZZ ANSWER THE QUESTION ​
mixas84 [53]
Its Displacement and Time for sure.
Thank You!
Pls mark Brainliest!
7 0
4 years ago
Find the total displacement of a mouse that travels 1.0 m north and then 0.8 m south.
andreev551 [17]
North: 1m
South: 0,8m

Direction:

1m>0,8m 

so mouse moved north

Distance:

1m-0,8m=0,2m

so mouse traveled 0,2m


Answer: The mouse moved 0,2m north.



"Non nobis Domine, non nobis, sed Nomini tuo da gloriam."



Regards M.Y.





3 0
3 years ago
Car A (1750 kg) is travelling due south and car B (1450 kg) is travelling due east. They reach the same intersection at the same
Contact [7]

Consider the east-west direction along x-axis and north-south direction along y-axis. In unit vector notation, velocities can be given as

\underset{V_{A}}{\rightarrow} = velocity of car A before collision = 0 i - V_{A} j

\underset{V_{B}}{\rightarrow} = velocity of car B before collision = V_{B} i + 0 j

\underset{V_{AB}}{\rightarrow} = velocity of combination after collision = (35.8 Cos31.6) i - (35.8 Sin31.6) j = 30.5 i - 18.8 j

M_{A} = mass of car A = 1750 kg

M_{B} = mass of car B = 1450 kg

Using conservation of momentum

M_{A}  \underset{V_{A}}{\rightarrow} + M_{B}  \underset{V_{B}}{\rightarrow} = (M_{A} + M_{B}) ( \underset{V_{AB}}{\rightarrow} )

(1750) (0 i - V_{A} j) + (1450) (V_{B} i + 0 j) = (1750 + 1450) (30.5 i - 18.8 j)

(1450) V_{B} i - (1750) V_{A} j = 97600 i - 60160 j

Comparing the coefficient of "i" and "j" both side

(1450) V_{B} = 97600    and - (1750) V_{A} = - 60160

V_{B} = 67.3 km/h        and  V_{A} = 34.4 km/



6 0
4 years ago
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