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2 years ago

15

Compared to the amount of radiation received from the Sun, about how much radiation does the surface of the earth receive from t

he atmosphere

1 answer:

babymother [125]2 years ago

5 0

Answer:

https://earthobservatory.nasa.gov/features/EnergyBalance#:~:text=The%20atmosphere%20absorbs%2023%20percent,surface%20radiates%20only%2012%20percent.

Explanation:

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using mass and distance, identify and compare the sun's and moon's contribution to the formation of tides on earth

vekshin1

Answer:

Based on its mass, the sun's gravitational attraction to the Earth is more than 177 times greater than that of the moon to the Earth.

7 0

3 years ago

6. Explain the difference between a neutral atom and an ion?

Deffense [45]

Answer:

a neutral atom doesn't have a charge and an ion does.

Explanation:

When atoms bond they can gain or lose electrons which gives them a charge.

anion = negatively charged ion

cation = positively charged ion

8 0

3 years ago

LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.

Svetradugi [14.3K]

Answer:

i) E = 269 [MJ] ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

$E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.$

$W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]$

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

$E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]$

8 0

3 years ago

Hi i do not have a question i just want to show u my cute doggo :)) ik everyone on here is probably stressed so hope this makes

Arisa [49]

Answer:

wooowwwwww

Explanation:

it's so so cuteeeeeeee

I love dogs tho:))

5 0

3 years ago

Read 2 more answers

Part A (4 pts) Consider light of wavelength λ = 670nm traveling in air. The light is incident at normal incidence upon a thin fi

s344n2d4d5 [400]

Answer:

Explanation:

On both sides of the film , the mediums have lower refractive index.

for interfering pattern from above , for constructive interference of reflected wave from both sides of the film , the condition is

2μt = ( 2n +1 ) λ / 2

μ is refractive index of film ,t is thickness of film λ is wavelength of light

n is order of fringe

for minimum thickness

n = 0

2μt = λ / 2

t = λ / 4μ

= 670 / 1.75 x 4

= 95.71 nm .

6 0

3 years ago