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finlep [7]
3 years ago
12

What would venus's "no greenhouse" temperature be if its clouds were more transparent, giving a reflectivity similar to earth's?

(b) what would the actual surface temperature now be if the greenhouse effect increased its temperature by the same number of degrees it does today?
Physics
2 answers:
LUCKY_DIMON [66]3 years ago
5 0
<span>Answer: Venus is nearer to the Sun, and gets relatively more power from it. Earth is 93 million miles from the Sun, by and large, while Venus is just 67.25 million (by and large) from the Sun. Since the force of the Sun's radiation diminishes with remove from it as 1 over r-squared, Venus gets (93/67.25) squared, or 1.91 times the power for each unit range that Earth gets, all things considered. Since the emanating temperature of a confined body in space fluctuates as the fourth-foundation of the power occurrence upon it, by the Stefan-Boltzmann law, the transmitting temperature of Venus ought to be the fourth-base of 1.91 = 1.18 times that of the Earth. Moreover, since the environmental weight differs as the temperature, the temperature at any given weight level in the Venusian air ought to be 1.18 times the temperature at that same weight level in the Earth climate, INDEPENDENT OF ANY INFRARED ABSORPTION in the air. Specifically, the found the middle value of temperature at 1000 millibars on Earth is around 15ÂşC = 288K, so the comparing temperature on Venus, WITHOUT ANY GREENHOUSE EFFECT, ought to be 1.18 times that, or 339K. Yet, this is only 66ÂşC, the temperature we really find there from the temperature and weight profiles for Venus. So there is no nursery impact. You have recently demonstrated that atmosphere science is totally wrong to suspect something. This is the embarrassment that such huge numbers of "specialists" in atmosphere science, and all the logical experts, won't confront. Tune in to the physicists that let you know there is no nursery impact; they know without going to the Venus information - and I am one of them. The proceeding with ineptitude on this fundamental point among such a significant number of researchers, for over a century, is stunning, and deplorable..</span>
Vlad1618 [11]3 years ago
3 0

Answer:

1. 29^{0} C

2.539^{0}C

Explanation:

Green house gases is known as the gases that educes the infrared radiation of the earth crust. If allows ultraviolet rays of the sun penetrate into the ozone layer(a protective covering in the earth's atmosphere.

Venus is known as the earth's twin because of it close proximity in size with the earth.

venus "no greenhouse" temperature will be 29^{0} C if its clouds were more transparent, giving a reflectivity similar to earth.

2, The actual surface temperature will be  539 deg C, if the green house effect increased . green house effect will actually increase the surface temperature of the venus.

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What is the centripetal acceleration of the earth as it travels around the sun when Earth has an orbital radius of 1.5 x 10^11 m
masya89 [10]

Answer: 0.0058 m/s^{2}

Explanation:

Centripetal acceleration a_{C} is calculated by the following equation:

a_{C}=\frac{V^{2}}{r}

Where:

V=29.7 \frac{km}{h} \frac{1000 m}{1 km}=29700 m/s is the Earth's orbital speed

r=1.5(10)^{11} m is the orbital radius

a_{C}=\frac{(29700 m/s)^{2}}{1.5(10)^{11} m}

a_{C}=0.0058 m/s^{2}

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Eating 2500 Cal every day a friend of mine maintains a stable weight of 70 kg. One day, after eating 3500 Cal, he decided to do
Kaylis [27]

Answer:

Explanation:

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4 0
3 years ago
A particle whose speed is 50 m/sec moves along the line from A(2,1) to B (9,25)
WINSTONCH [101]

First, calculate for the distance between the given points A and B by using the equation,

<span>                                                D = sqrt ((x2 – x1)2 + (y2 – y1)2)</span>

 

Substitute the known values:

<span>                                                D = sqrt((9 – 2)2 + (25 – 1)2)</span>

<span>                                                D = 25 m</span>

 

I assume the unknown here is the time it would require for the particle to move from point A to B. This can be answered by dividing the calculated distance by the speed given above.

<span>                                                t = (25 m)/ (50 m/s) = 0.5 s</span>

 

<span>Thus, it will take 0.5s for the particle to complete the route. </span>

3 0
3 years ago
A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is
BlackZzzverrR [31]

Answer:

the penny loses contact at the piston's highest point.

f = 2.5 Hz

Explanation:

Concepts and Principles  

1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by  

∑F = ma                                                          (1)  

2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:  

a_max = -w^2A                                                (2)  

3- The angular frequency w of a wave is related to the frequency f by:  

w = 2π f                                                              (3)  

Given Data  

- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)=  0.04 m.  

- The frequency of oscillation of the piston is steadily increased.

Required Data

<em>In part (a), we are asked to determine the point at which the penny first loses contact with the piston.  </em>

<em>In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle.  </em>

Solution  

(a)  

The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.  

figure 1 is attached

Apply Newton's second law from Equation (1) in the vertical direction to the penny:  

∑F_y -mg= ma        

Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So  

0 = m(g + a)

a = —g  

Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.

(b)  

The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):  

a = —w^2A  

where a = —g at the highest point. So  

g = w^2A  

Solve for w:  

w =√g/A

Substitute for w from Equation (3):

2πf =  √g/A

Solve for f :  

f = 1/2π√g/A

Substitute numerical values:  

f = 1/2π√9.8 m/s^2/0.04

f = 2.5 Hz

6 0
3 years ago
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