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OlgaM077 [116]
4 years ago
13

The Earth's oceans act as a climate moderator. Oceans do this in many ways, including ALL BUT one method listed here. That is A)

evenly distributing solar radiation. B) absorbing and releasing solar energy. C) absorbing huge amounts of carbon dioxide. D) providing moisture to Earth's atmosphere.
Physics
2 answers:
mamaluj [8]4 years ago
8 0

Answer:

A

Explanation:

evenly distributing solar radiation.

LUCKY_DIMON [66]4 years ago
3 0

Answer:

Option D is correct that is providing moisture to Earth's atmosphere.

Step-By-Sep explanation:

The Earth's  oceans act as a climate  Moderator

The water from oceans gets evaporated which eventually provide moisture to the Earth's atmosphere .Moisture Creates humidity in the atmosphere which increases the temperature and causes greenhouse affect eventually causes global warming.

Therefore, option D is correct that is providing moisture to Earth's atmosphere.

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Is fire actually hot and why do we call it hot; is it a classification for when something burns you.
rodikova [14]
Yes it is classified as hot
6 0
4 years ago
An example of convection and conduction within a BBQ. I also accept diagrams but please write honestly or I will report.
qwelly [4]
Conduction could be the way in which heat is transferred through a sausage to get it up to cooked temperature. convection could be the air around the barbecue and the smell of the barbecue being carried on thermal convection currents. hope this is honest enough
7 0
3 years ago
A boy throws a ball straight up with a speed of 21.5 m/s. The ball has a mass of 0.19 kg. How much gravitational potential energ
astra-53 [7]

Answer:

Explanation:

The equation fo potential energy is PE = mgh, where m is the mass of the ball, g is the pull of gravity (constant at 9.8), and h is the max height of the ball. What we do not have here is that height. We need to first solve for it using one-dimensional equations. What we have to know above all else, is that the final velocity of an object at its max height is always 0. That allows us to use the equation

v_f=v_0+at where vf is the final velocity and v0 is the initial velocity. We will find out how long it takes for the object to reach that max height first and then use that time to find out what that max height is. Baby steps here...

0 = 21.5 + (-9.8)t and

-21.5 = -9.8t so

t = 2.19 seconds (Keep in mind that if I used the rules correctly for sig fig's, the answer you SHOULD get is not one shown, so I had to adjust the sig fig's and break the rules. But you know what they say about rules...)

Now we will use that time to find out the max height of the object in the equation

Δx = v_0t+\frac{1}{2}at^2 and filling in:

Δx = 21.5(2.19)+\frac{1}{2}(-9.8)(2.19)^2 which simplifies down a bit to

Δx = 47.1 - 23.5 so

Δx = 23.6 meters.

Now we can plug that in to the PE equation to find the PE of the object:

PE = (.19)(9.8)(23.6) so

PE = 43.9 J

5 0
3 years ago
A football player, with a mass of 69.0 kg, slides on the ground after being knocked down. At the start of the slide, the player
White raven [17]

Answer:

(a) -472.305  J

(b) 1 m

Explanation:

(a)

Change in mechanical energy equals change in kinetic energy

Kinetic energy is given by0.5mv^{2}

Initial kinetic energy is 0.5\times 69\times 3.7^{2}=472.305 J

Since he finally comes to rest, final kinetic energy is zero because the final velocity is zero

Change in kinetic energy is given by final kinetic energy- initial kinetic energy hence

0-472.305  J=-472.305  J

(b)

From fundamental kinematic equation

v^{2}=u^{2}+2as

Where v and u are final and initial velocities respectively, a is acceleration, s is distance

Making s the subject we obtain

s=\frac {v^{2}-u^{2}}{-2a} but a=\mu g hence

s=\frac {v^{2}-u^{2}}{-2\mu g}=\frac {0^{2}-3.7^{2}}{-2*0.7*9.81}=0.996796272\approx 1 m

7 0
3 years ago
A basketball player standing up with the hoop launches the ball straight up with an initial velocity of v_o = 3.75 m/s from 2.5
denis23 [38]

Answer:

a) The maximum height the ball will achieve above the launch point is 0.2 m.

b) The minimum velocity with which the ball must be launched is 4.43 m/s or 0.174 in/ms.

Explanation:

a)

For the height reached, we use 3rd equation of motion:

2gh = Vf² - Vo²

Here,

Vo = 3.75 m/s

Vf =  0m/s, since ball stops at the highest point

g = -9.8 m/s² (negative sign for upward motion)

h = maximum height reached by ball

therefore, eqn becomes:

2(-9.8m/s²)(h) = (0 m/s)² - (3.75 m/s²)²

<u>h = 0.2 m</u>

b)

To find out the initial speed to reach the hoop at height of 3.5 m, we again use 3rd eqn. of motion with h= 3.5 m - 2.5m = 1 m (taking launch point as reference), and Vo as unknown:

2(-9.8m/s²)(1 m) = (0 m/s)² - (Vo)²

(Vo)² = 19.6 m²/s²

Vo = √19.6 m²/s²

<u>Vo = 4.43 m/s</u>

Vo = (4.43 m/s)(1 s/1000 ms)(39.37 in/1 m)

<u>Vo = 0.174 in/ms</u>

<u />

6 0
3 years ago
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