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3 years ago

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The temperature at the surface of the Sun is approximately 5,200 K, and the temperature at the surface of the Earth is approxima

tely 295 K. What entropy change of the Universe occurs when 4.50 103 J of energy is transferred by radiation from the Sun to the Earth

1 answer:

Sever21 [200]3 years ago

5 0

To solve this problem, we will calculate the entropy for both cases, remembering that the concept of entropy is the relationship between the heat released / gained and the temperature. After calculating the entropy in the sun and on the earth we will find the difference between the two. So that,

Entropy at Sun

$S_1 = \frac{Q}{T_1}$

$Q= 4500 J$

$T_1 = 5200 K$

Replacing,

$S_1 =\frac{4500}{5200}$

$S_1 = 0.86 J \cdot K^{-1}$

The entropy at Earth,

$S_2 = \frac{Q}{T_2}$

The values are,

$T_2 = 290K$

$Q = 4500J$

Replacing at the equation,

$S_2 = \frac{4500}{290}$

$S_2 = 15.51J \cdot K^{-1}$

Then the total change in entropy will be,

$\Delta S = S_2 -S_1$

$\Delta S = 15.51-0.86$

$\Delta S = 14.65 J \cdot K^{-1}$

Therefore the entropy change is $14.65J \cdot K^{-1}$

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A car is going 3 m/s, then it speeds up for 4 seconds. If it has an acceleration of 5m/s^2, find its final velocity. Which equat

andreyandreev [35.5K]

Well, since the question GIVES you the initial velocity, the acceleration, and the time, and ASKS for the final velocity, you'd be smart to find an equation that USES the initial velocity, the acceleration, and the time, and FINDS the final velocity.

Have a look at equation B ..... vf = vi + a*t . That's pretty durn close !

vf = (initial velocity) + (acceleration)*(time)

vf = (3 m/s) + (5 m/s²)*(4 sec)

vf = (3 m/s) + (20 m/s)

vf = 23 m/s

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3 years ago

Hy would you expect sodium (Na) to react strongly with chlorine (Cl)?

erastova [34]

Chlorine needs to gain one electron and sodium needs to lose one electron

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3 years ago

Everyday lifestyle choices affect your physical fitness. Please select the best answer from the choices provided. T F

amid [387]

Answer:
True

Explanation:
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The everyday lifestyle affects this physical fitness, in case you are not used used to "moving" in your daily life, your body will be sort of lazy and lack the power and ability to carry on with the daily occupation as well as sports.
In case your daily lifestyle involved motion, your body will be used to doing activities which will help improve your physical fitness.

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3 years ago

Read 2 more answers

How many subatomic particles are in the element moscovium? Will give brainliest!!

Alja [10]

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3 years ago

Interactive Solution 6.39 presents a model for solving this problem. A slingshot fires a pebble from the top of a building at a

mariarad [96]

(a) 29.8 m/s

To solve this problem, we start by analyze the vertical motion first. This is a free fall motion, so we can use the following suvat equation:

$v_y^2 - u_y^2 = 2as$

where, taking upward as positive direction:

$v_y$ is the final vertical velocity

$u_y = 0$ is the initial vertical velocity (zero because the pebble is launched horizontally)

$a=g=-9.8 m/s^2$ is the acceleration of gravity

s = -25.0 m is the displacement

Solving for vy,

$v_y = \sqrt{u^2+2as}=\sqrt{0+2(-9.8)(-25)}=-22.1 m/s$ (downward, so we take the negative solution)

The pebble also have a horizontal component of the velocity, which remains constant during the whole motion, so it is

$v_x = 20.0 m/s$

So, the final speed of the pebble as it strikes the ground is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{20.0^2+(-22.1)^2}=29.8 m/s$

(b) 29.8 m/s

In this case, the pebble is launched straight up, so its initial vertical velocity is

$u_y = 20.0 m/s$

So we can find the final vertical velocity using the same suvat equation as before:

$v_y^2 - u_y^2 = 2as$

$v_y = \sqrt{u^2+2as}=\sqrt{(20.0)^2+2(-9.8)(-25)}=-29.8 m/s$ (downward, so we take the negative solution)

The horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

(c) 29.8 m/s

This case is similarly to the previous one: the only difference here is that the pebble is launched straight down instead than up, therefore

$u_y = -20.0 m/s$

Using again the same suvat equation:

$v_y^2 - u_y^2 = 2as$

$v_y = \sqrt{u^2+2as}=\sqrt{(-20.0)^2+2(-9.8)(-25)}=-29.8 m/s$ (downward, so we take the negative solution)

As before, the horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

We notice that the final value of the speed is always the same in all the three parts, so it does not depend on the direction of launching. This is due to the law of conservation of energy: in fact, the initial mechanical energy of the pebble (kinetic+potential) is the same in all three cases (because the height h does not change, and the speed v does not change either), and the kinetic energy gained during the fall is also the same (since the pebble falls the same distance in all 3 cases), therefore the final speed must also be the same.

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3 years ago