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3 years ago

The ocean’s surface temperatures vary with time of year and a. latitude. c. seasons. b. longitude. d. moon phases.

Physics

2 answers:

Natali5045456 [20]3 years ago

5 0

answer is A your welcome

nignag [31]3 years ago

4 0

Answer: a. Latitude

Explanation:

The average temperature of ocean's surface is about 17°C. The temperature varies with the seasons and location of the ocean. It means the time of the year -whether it is winters , summers, autumn or spring and location - in which zone it lies.

In winters the temperature dips down. The arctic ocean and Antarctic ocean have extremely low temperatures. On the other ocean lying near equator have higher temperature.

Thus, the ocean’s surface temperatures vary with time of year and<u> a. latitude. </u>

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What is the highest frequency wave? The lowest frequency wave?

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Answer:

Gamma rays have the highest energies, the shortest wavelengths, and the highest frequencies. Radio waves, on the other hand, have the lowest energies, longest wavelengths, and lowest frequencies of any type of EM radiation.

Explanation:

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3 years ago

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The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.2

sergejj [24]

Answer:

The car strikes the tree with a final speed of 4.165 m/s

The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m

Explanation:

First we need to calculate the initial speed $V_{0}$

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62.5m=V_{0} *4.15s+\frac{1}{2} *-5.25\frac{m}{s^{2} } *(4.15^{2} )\\V_{0}=25.953\frac{m}{s}$

Once we have the initial speed, we can isolate the final speed from following equation:

$V_{f} =V_{0} +a*t$ $V_{f}= 4.165 \frac{m}{s}$

Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.

To do that, we replace 62 m in the first formula, as follows:

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62m=25.953\frac{m}{s}*4.15s+\frac{1}{2} *-a\frac{m}{s^{2} } *(4.15^{2} )\\a=-5.19\frac{m}{s^{2} }$

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3 years ago

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A child on a swing sweeps out a distance of 45 ft on the first pass. If she is allowed to continue swinging until she stops, an

earnstyle [38]

Answer:

d = 90 ft

Explanation:

Here in each swing the distance sweeps by the swing is half of the initial distance that it will move

So here we can say that total distance in whole motion is given as

$d = 45 + \frac{45}{2} + \frac{45}{4} + \frac{45}{8}..........$

since it is half of the distance that it will move in each swing so it would be a geometric progression with common ratio of 1/2

so sum of such GP is given by the formula

$S = \frac{a}{1 - r}$

$d = \frac{45}{1 - \frac{1}{2}}$

$d = 90 ft$

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3 years ago

A less massive moving object has an elastic collision with a more massive object that is not moving. Compare the initial velocit

stepladder [879]

Assume that the small-massed particle is $m$ and the heavier mass particle is $M$ .

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Now, there are 2 solutions but, one of them is useless to this question's main point so I excluded that point. Ask me in the comments if you want the excluded solution too.

$v_{m} = v\frac{m - M}{m + M}\\\\\\v_{M} = v\frac{m + M}{m + M}\\$