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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 a. latitude.

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The sidereal period of the moon around the Earth is 27.3 days. Suppose a satellite were placed in Earth orbit, halfway between E

icang [17]

Answer: 9.7 days

Explanation:

Applying Kepler's 3rd law, we can write the following proportion:

(Tm)² / (dem)³ = (Tsat)² / (dem/2)³

(As the satellite is placed in an orbit halfway between Erth's center and the moon's orbit).

Simplifyng common terms, and solving for Tsat, we have:

Tsat = √((27.3)²/8) = 9.7 days

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3 years ago

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude

Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

$\omega_f = \omega_i + \alpha t$

where

$\omega_i = 0.300 rev/s$ is the initial angular velocity

$\alpha = 0.895 rev/s^2$ is the angular acceleration

Substituting t = 0.200 s, we find

$\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s$

Let's now convert it into rad/s:

$\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s$

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

$r=\frac{0.800}{2}=0.400 m$

And so now we can find the tangential speed at t = 0.200 s:

$v=\omega_f r =(3.01)(0.400)=1.2 m/s$

2) $2.25 m/s^2$

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

$a_t = \alpha r$

where $\alpha$ is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

$\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2$

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

$a_t = (5.62)(0.400)=2.25 m/s^2$

3) $3.6 m/s^2$

The centripetal acceleration is given by

$a=\frac{v^2}{r}$

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

$a=\frac{1.2^2}{0.400}=3.6 m/s^2$

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3 years ago

Is work being done on a barbell when a weight lifter is holding the barbell over his head?

AVprozaik [17]

I don’t see any picture sorry I wish I can help tho

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3 years ago

________ is the average kinetic energy of each atom. radiation temperature potential

SSSSS [86.1K]

It could be radiation because radiation means the emission of energy as electromagnetic wave, but potential fits it best

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3 years ago

A baseball pitcher loosens up his pitching arm. He tosses a 0.20-kg ball using only the rotation of his forearm, 0.28 m in lengt

aev [14]

Answer:

Moment of Inertia, I = 0.016 kgm²

Explanation:

Mass of the ball, m = 0.20 kg

Length of the pitcher's arm, l = 0.28

Radius of the circular arc, r = 0.28 m

Moment of Inertia is given by the formula:

I = mr²

I = 0.20 * 0.28²

I = 0.20 * 0.0784

I = 0.01568

I = 0.016 kgm²

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3 years ago