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3 years ago

How does the Earth release energy back into the atmosphere?

1 answer:

8 0

In general, the Earth releases energy back to the atmosphere through reflection, evaporation, and radiation. The Earth gets energy from the sunlight, part of which it absorbs, while part it reflects backwards, thus giving energy to the atmosphere. Also, the heating up of the Earth by the absorbed sunlight, radiates back in the lower layers of the atmosphere, again giving back energy to it. The water vapor is another way in which the Earth gives back energy tot he atmosphere as through the evaporation, the water vapor gets into the lower parts of the atmosphere and gives energy to it.

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A fish swimming in a horizontal plane has velocity i = (4.00 î + 1.00 ĵ) m/s at a point in the ocean where the position relative

Liula [17]

Question is missing. Found on google:

a) What are the components of the acceleration of the fish?

(b) What is the direction of its acceleration with respect to unit vector î?

(a) $(0.73, -0.47) m/s^2$

The initial velocity of the fish is

$u=(4.00 i + 1.00 j) m/s$

while the final velocity is

$v=(15.0 i - 6.00 j) m/s$

Initial and final velocity are related by the following suvat equation:

$v=u+at$

where

a is the acceleration

t is the time

The time in this case is t = 15.0 s, so we can use the previous equation to find the acceleration, separating the components:

$v_x = u_x + a_x t\\a_x = \frac{v_x-u_x}{t}=\frac{15.0-4.00}{15.0}=0.73 m/s^2$

$v_y = u_y + a_y t\\a_y = \frac{v_y-u_y}{t}=\frac{-6.00-1.00}{15.0}=-0.47 m/s^2$

(b) $-32.8^{\circ}$

The direction of the acceleration vector with respect to i can be found by using the formula

$\theta = tan^{-1}(\frac{a_y}{a_x})$

where

$a_x$ is the horizontal component of the acceleration

$a_y$ is the vertical component of the acceleration

From part a), we have

$a_x = 0.73 m/s^2$

$a_y = -0.47 m/s^2$

Substituting,

$\theta = tan^{-1}(\frac{-0.47}{0.73})=-32.8^{\circ}$

(c) $r=(460.5 i - 185.1 j )m$

The initial position of the fish is

$r_0 = (12.0 i -3.60 j) m$

The generic position r at time t is given by

$r= r_0 + ut + \frac{1}{2}at^2$

where

$u=(4.00 i + 1.00 j) m/s$ is the initial velocity

$a=(0.73 i -0.47 j) m/s^2$ is the acceleration

Substituting t = 30.0 s, we find the final position of the fish. Separating each component:

$r_x =12.0 + (4.00)(30) + \frac{1}{2}(0.73)(30)^2=460.5 m\\r_y = -3.60 + (1.00)(30) + \frac{1}{2}(-0.47)(30)^2=-185.1 m$

So the final position is

$r=(460.5 i - 185.1 j )m$

4 0

3 years ago

Will pushing on a car always change the cars mechanical energy?what must happen for the cars kinetic energy to increase?

riadik2000 [5.3K]

1) Pushing on a car will not always change the car's mechanical energy, but it does change yours since your physically using most of your energy to push the car.

2) In order for a cars kinetic energy to increase, the car has to be in motion when it is going down a hill, because when you go down hills and objects like that; you tend to increase in kinetic energy.

7 0

3 years ago

Read 2 more answers

Please answer this question

steposvetlana [31]

Hope you understand

Have a nice day