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MrRa [10]
4 years ago
14

Part of the Sun's energy that reaches the Earth's surface is absorbed by land and water as heat. The Earth's surface then releas

es heat back to the atmosphere.
Which of the following is true about this solar energy that is absorbed and released by Earth's surface as heat?

Question 9 options:

Much of the heat is trapped at the top of Earth's atmosphere to protect Earth from ultraviolet radiation.


All of this heat escapes the Earth's atmosphere, which keeps the planet cool.


Much of the heat is trapped low in the atmosphere and is a major factor in determining Earth's climate.


All of this heat escapes the Earth's atmosphere, which keeps the planet hot.
Physics
2 answers:
Elina [12.6K]4 years ago
8 0

Answer:

option c is correct

Explanation:

hope you guys get it right

BARSIC [14]4 years ago
3 0
Option C .. as more amount of energy is trapped at the lower part of atmosphere
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Increasing the concentration and increasing the temperature both increase reaction rates based on which physical phenomenon? 1.
vekshin1
You can deal with this question using the collision theory, which states that the chemical reactions occur when particles collide with enough energy to reach the Activation Energy.

The velocity of the particles is related with the temperature. At higher temperatures,  higher velocities and higher frequency of collisions.

Also, at higher concentration (more particles is a same volume) the number of collisions will increase.

Then, hIgher concentration and higher temperature will increase the frequency of the molecular collisions..

Then, the answer is the proposal #1: "<span>increasing the frequency of molecular collisions"</span>
 

5 0
3 years ago
Read 2 more answers
The cart now moves toward the right with an acceleration toward the right of 2.50 m/s2. What does spring scale Fz read? Show you
katrin2010 [14]

Complete Question

The  complete question is shown on the first uploaded image

Answer:

The spring scale F_2 reads  F_2 =  2.4225 \ N

Explanation:

From the question we are told that

      The first force is  F_1  =  10.5 \ N

      The acceleration by which the cart moves to the right is  a = 2.50  \ m/s^2

      The mass of the cart is  m  = 3.231  kg

       

Generally the net force on the cart is  

       F_{net} = F_1 -  F_2

This net force is mathematically represented as

      F_{net} =  m * a

So  

        m*  a =  10 - F_2

        F_2 =  10.5 -  2.5 (3.231)

        F_2 =  2.4225 \ N

 

3 0
3 years ago
PLEASE HELP!!!!
Aleks [24]
The answer is step by step 65
4 0
3 years ago
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The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles
miss Akunina [59]

Formula of the gravitational force between two particles:

F=G\frac{m_1 m_2}{r^2}

where

G=6.67 \cdot 10^{-11} Nm^2 kg^{-2} is the gravitational constant

m1 and m2 are the masses of the two particles

r is their distance


(a) particle A

The gravitational force exerted by particle B on particle A is

F_B=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=5.14 \cdot 10^{-5} N to the right

The gravitational force exerted by particle C on particle A is

F_C=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=5.6 \cdot 10^{-6} N to the right

So the net gravitational force on particle A is

F_A = F_B + F_C =5.14 \cdot 10^{-5} N+5.6 \cdot 10^{-6} N=5.7 \cdot 10^{-5} N to the right


(b) Particle B

The gravitational force exerted by particle A on particle B is

F_A=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=-5.14 \cdot 10^{-5} N to the left

The gravitational force exerted by particle C on particle B is

F_C=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=8.41 \cdot 10^{-5} N to the right

So the net gravitational force on particle B is

F_B = F_A + F_C =-5.14 \cdot 10^{-5} N+8.41 \cdot 10^{-5} N=3.27 \cdot 10^{-5} N to the right


(c) Particle C

The gravitational force exerted by particle A on particle C is

F_A=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=-5.6 \cdot 10^{-6} N to the left

The gravitational force exerted by particle B on particle C is

F_B=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=-8.41 \cdot 10^{-5} N to the left

So the net gravitational force on particle C is

F_C = F_B + F_A =-8.41 \cdot 10^{-5} N-5.6 \cdot 10^{-6} N=-8.97 \cdot 10^{-5} N to the left



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