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vlabodo [156]
3 years ago
6

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
Physics
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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Explanation:

We start by using the conservation law of energy:

\Delta{K} + \Delta{U} = 0

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\dfrac{1}{2}mv^2 - G\dfrac{mM}{r} = 0

Simplifying the above equation, we get

v^2 = 2G\dfrac{M}{r}

We can rewrite this as

v^2 = 2\left(G\dfrac{M}{r^2}\right)r

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v^2 = 2gr

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3 years ago
What is the only component of scalar quantities?
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Answer:

Scalar quantities have a size or magnitude only and need no other information to specify them. Thus, 10 cm, 50 sec, 7 litres and 3 kg are all examples of scalar quantities.

Explanation:

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3 years ago
A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia
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Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

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What is the acceleration of a car that goes from 0 to 20m/s in 7 seconds?
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Answer:

2m/s^2

Explanation:

Clculate the acceleration:

V = u +at

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a = 20m//10s

a = 2m/s²

From the data given , it is not possible to calculate the displacement , because no direction of motion is given

But it is possible to calculate the distance travelled

Distance = ut + ½ *a*t²

distance = 0 + ½ * 2m/s * 10²s

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