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3 years ago

There is a balance of heat coming into and going out of earth. This is known as the?

Physics

2 answers:

valkas [14]3 years ago

3 0

Earths energy budget

kobusy [5.1K]3 years ago

3 0

Answer:

Earth's energy budget states balance between the energy that the Earth collects from Sun and energy that Earth radiates back into earth orbit after distribution through out the components of Earth's climate system, thereby powering the so-called heat engine of Earth.

Explanation:

This particular equilibrium condition is called radiative equilibrium. Around 29% of solar energy reaching the top of atmosphere is transmitted back into space by clouds, dust particles and light surfaces on the ground.

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Assume that when you stretch your torso vertically as much as you can, your center of mass is 1.0 m above the floor. The maximum

Elenna [48]

1) 0.77 m

2) 0.23 m

Explanation:

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

$\frac{1}{2}mv^2 = mgh$

where

m is the mass of the man

v is the speed after jumping

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 2.0 - 1.0 = 1.0 m is the change in height

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s$

In the acceleration phase, we know that the initial velocity is

$u=0$

And the force exerted on the floor is 2.3 times the gravitational force, so

$F=2.3 mg$

This means the net force on you is

$F_{net} = F-mg=2.3mg-mg=1.3 mg$

because we have to consider the force of gravity acting downward.

So the acceleration of the man is

$a=\frac{F_{net}}{m}=\frac{1.3mg}{m}=1.3g$

Now we can use the following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

$v^2-u^2=2as$

where s is the quantity we want to find. Solving for s,

$s=\frac{v^2-u^2}{2a}=\frac{4.43^2-0}{2(1.3g)}=0.77 m$

At the beginning, we are told that the height of the center of mass above the floor is

h = 1.0 m

During the acceleration phase and the crouch, the height of the center of mass of the body decreases by

$\Delta h = -0.77 m$

This means that the lowest point reached by the center of mass above the floor during the crouch is

$h'=h+\Delta h = 1.0 - 0.77 = 0.23 m$

This value seems unpractical, since it is not really easy to crouch until having the center of mass 0.23 m above the ground.

3 0

3 years ago

On a velocity vs. time graph, what does it mean when the line crosses over the x-axis?

Phantasy [73]

On a velocity - time graph, if the line crosses the x - axis it depicts that the object has started moving in the opposite direction.

8 0

2 years ago

Why does it take more time for a larger sample of water to freeze?

Stels [109]

Answer:The greater the amount of water that there is it will take longer for the water to freeze because more heat has to be dissipated into the environment

Explanation:

3 0

2 years ago

Read 2 more answers

At a particular instant, a proton at the origin has velocity < 5e4, -2e4, 0> m/s. You need to calculate the magnetic field

vesna_86 [32]

Answer:

$9.7\times 10^{-5} T$