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2 years ago

Similarities and differences between the north and pacific ocean currents

1 answer:

4 0

The Pacific is loosely shaped like a triangle, opening widely to the south but barely at all to the north, while the Atlantic is shaped like an hourglass with the choke point located very loosely at the equator (somewhat south of it in the west).

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Look at the diagram of the moon and three different paths that it could take. A planet with a moon in orbit. The moon has arrows

marysya [2.9K]

Answer:

is 3 and 2

Explanation: the firth one is 3 and the 2

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2 years ago

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Which of the outer planets would you most likely visit? Why? What would you see on your visit?

Tanzania [10]

I would like to visit Pluto because i want to see what a Dwarf planet would look like, i would like to see what kind of minerals are in the planet its self..

Brainliest answer?

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3 years ago

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An incident ray that passes through the vertex of a convex lens:

WINSTONCH [101]

The answer is refracts parallel to the axis of the lens

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2 years ago

Two trains A and B of length 400 m each are moving on two parallel tracks with

Vinvika [58]

Answer:

The initial distance between the trains is 1450 m. 

Explanation:

In the question two trains are of equal length 400 m and moves at a uniform speed of 72 km/h. train A is moving ahead of train B. If the train B has to overtake train A it should accelerate.

Train B’s acceleration is $1m/s^2$ and it accelerated for 50 seconds.

 $a=1 m/s^2$ 

t=50 s 

initial speed u=72km/h 

we have to convert this speed into m/s 

 $u=72 \times 5/18=20 m/s$ 

Distance covered in accelerating phase $S=ut+1/2 at^2$ 

 $=20 \times 50+1/2 \times 1 \times 50^2$ 

 $=1000+1250=2250 m$ 

If a train is just behind another, the distance covered by the train located behind during overtaking phase will be equal to the sum of the lengths of the trains.

Here length of train A+length of train $B=400+400=800 m$ 

Hence the initial distance between the trains = $2250-800=1450 m$ 

6 0

3 years ago

A ball is hit with a paddle, causing it to travel straight upward. It takes 2.90 s for the ball to reach its maximum height afte

zmey [24]

Answer:

A. 28.42 m/s

B. 41.21 m

Explanation:

From the question given above, the following data were obtained:

Time (t) to reach the maximum height = 2.90 s

Initial velocity (u) =?

Maximum height (h) =?

A. Determination of the initial velocity of the ball.

Time (t) to reach the maximum height = 2.90 s

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =?

v = u – gt (since the ball is going against gravity)

0 = u – (9.8 × 2.9)

0 = u – 28.42

Collect like terms

0 + 28.42 = u

u = 28.42 m/s

Thus, the initial velocity of the ball is 28.42 m/s

B. Determination of the maximum height reached by the ball.

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) = 28.42 m/s

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 28.42² – (2 × 9.8 × h)

0 = 807.6964 – 19.6h

Collect like terms

0 – 807.6964 = – 19.6h

– 807.6964 = – 19.6h

Divide both side by – 19.6

h = – 807.6964 / – 19.6

h = 41.21 m

Thus, the maximum height reached by the ball is 41.21 m

5 0

3 years ago