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3 years ago

12

Who was the first man on planet Jupiter

1 answer:

melamori03 [73]3 years ago

7 0

No one has been there yet

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Based on the simple blackbody radiation model described in class, answer the following question. The planets Mars and Venus have

Lera25 [3.4K]

Answer:

The extent of greenhouse effect on mars is $G_m = 87 K$

Explanation:

From the question we are told that

The albedo value of Mars is $A_1 = 0.15$

The albedo value of Mars is $A_2 = 0.15$

The surface temperature of Mars is $T_1 = 220 K$

The surface temperature of Venus is $T_2 = 700 K$

The distance of Mars from the sun is $d_m = 2.28*10^8 \ km = 2.28*10^8* 1000 = 2.28*10^{11} \ m$

The distance of Venus from sun is $d_v = 1.08 *10^{8} \ km = 1.08 *10^{8} * 1000 = 1.08 *10^{11} \ m$

The radius of the sun is $R = 7*10^{8} \ m$

The energy flux is $E = 6.28 * 10^{7} W/m^2$

The solar constant for Mars is mathematically represented as

$T = [\frac{E R^2 (1- A_1)}{\sigma d_m} ]$

Where $\sigma$ is the Stefan's constant with a value $\sigma = 5.6*10^{-8} \ Wm^{-2} K^{-4}$

So substituting values

$T = \frac{6.28 *10^{7} * (7*10^8)^2 * (1-0.15)}{(5.67 *10^{-8}) * (2.28 *10^{11})^2)}$

$T = 307K$

So the greenhouse effect on Mars is

$G_m = T - T_1$

$G_m = 307 - 220$

$G_m = 87 K$

3 0

4 years ago

Does earths surface heat up the same everywhere

melomori [17]

the correct answer is no ;)

4 0

3 years ago

Read 2 more answers

If the boy and his bike actually had a mass of 40 kg instead of 50 kg how much force would it take for him to accelerate at 0.8

klio [65]

Answer:

F = 32 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = forces applied [N] (units of Newtons)

m = mass = 40 [kg]

a = acceleration = 0.8 [m/s²]

Now replacing:

$F=m*a\\F = 40*0.8\\F = 32 [N]$

8 0

3 years ago

Important parts of stydi g physics

sladkih [1.3K]

Answer:

put your question in proper way' i hope you understand.

Explanation:

7 0

2 years ago

an object that is 8 cm tall is located 2.5 m in front of a plane mirror. The image formed by the mirror appears to be. Find di a

UkoKoshka [18]

The image is at 2.5 m behind the mirror (virtual image)

Explanation:

We can solve the problem by using the mirror equation:

$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$

where

f is the focal length of the mirror

$d_o$ is the distance between the object and the mirror

$d_i$ is the distance between the image and the mirror

For a plane mirror, the focal length is infinite:

$f=\infty$

And in this problem, the distance of the object from the mirror is

$d_o = 2.5 m$

Substituting and solving for $d_i$ , we find the location of the image:

$0=\frac{1}{d_o}+\frac{1}{d_i}\\d_i = -d_o = -2.5 m$

And the negative sign means that the image is located behind the mirror, so it is virtual.

Learn more about reflection and refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

6 0

3 years ago