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Lemur [1.5K]
3 years ago
12

The temperature at the surface of the Sun is approximately 5,300 K, and the temperature at the surface of the Earth is approxima

tely 293 K. What entropy change of the Universe occurs when 6.00 103 J of energy is transferred by radiation from the Sun to the Earth?
Physics
1 answer:
N76 [4]3 years ago
4 0

Answer:

The entropy change of the Universe that occurs is 19.346 J/K

Explanation:

Given;

temperature of the sun, T_s = 5,300 K

temperature of the Earth, T_E = 293 K

radiation energy transferred by the sun to the earth, E = 6000 J

The sun loses Q of heat and therefore decreases its entropy by the amount

\delta S_{sun} = \frac{-Q}{T_s}

The earth gains Q of heat and therefore increases its entropy by the amount

\delta S_{Earth} = \frac{-Q}{T_E}

The total entropy change is:

\delta S_{Earth} + \delta S_{sun} = \frac{Q}{T_E} -\frac{Q}{T_S} \\\\                                                      = Q(\frac{1}{T_E} -\frac{1}{T_S} )\\\\= 6000(\frac{1}{293} -\frac{1}{5300} )\\\\=6000(0.0032243)\\\\= 19.346 \ J/K

Therefore, the entropy change of the Universe that occurs is 19.346 J/K

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Dos masas de 8kg es tan unidas en el extremo de una varilla de aluminio de 400mm de longitud. La varilla está sostenida en su pa
enyata [817]

Answer:

The maximum frequency of revolution is 3.6 Hz.

Explanation:

Given that,

Mass = 8 kg

Distance = 400 mm

Tension = 800 N

We need to calculate the velocity

Using centripetal force

F=\dfrac{mv^2}{r}

Where, F= tension

m = mass

v= velocity

r = radius of circle

Put the value into the formula

800=\dfrac{8\times v^2}{200\times10^{-3}}

v^2=\sqrt{\dfrac{800\times200\times10^{-3}}{8}}

v=4.47\ m/s

We need to calculate the maximum frequency of revolution

Using formula of frequency

f=\dfrac{v}{2\pi r}

Put the value into the formula

f=\dfrac{4.47}{2\pi\times200\times10^{-3}}

f=3.6\ Hz

Hence, The maximum frequency of revolution is 3.6 Hz.

7 0
3 years ago
What is the interaction between surface water and groundwater in a watershed?
Rashid [163]

Answer:

Groundwater occasionally discharges into surface water and then, they flow together as a body of water in a watershed.

Explanation:

According to www.mbgnet.net A watershed describes an area of land that contains a common set of streams and rivers that all drain into a single larger body of water, such as a larger river, a lake or an ocean.

Therefore, when groundwater discharges into a body of surface water, for example a stream, the stream just like several other streams in a watershed would flow into a larger body of water.

3 0
4 years ago
The distance, measured from one edge of a black band to the same edge of the next band, is 5.0 cm. What additional information i
elena55 [62]

In order to find the average speed we will use the formula

v_{avg} = \frac{distance}{time}

now here we know that distance from one edge to other edge is given as

d = 5.00 cm

now in order to find the average speed as per the formula above we need to know the time taken from one edge to other edge

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3 years ago
Acceleration is defined as the rate of change for which characteristic?
VARVARA [1.3K]
Acceleration is defined as the rate of change of velocity.
8 0
3 years ago
Earth has a surface area of 197 million square miles. Convert this area into each of the following units.
slamgirl [31]

Answers:

a) 510,225,121.4 km^{2}

We know the Earth's surface area in square miles is:

197000000 mi^{2}

On the other hand, we know 1 mi=1.60934 km.

Then:

197000000 mi^{2} \frac{(1.60934 km)^{2}}{1 mi^{2}}=510,225,121.4 km^{2}

b) 510.225 Mm^{2}

In this part, we can work with the obtained value in part a:

510,225,121.4 km^{2}

And knowing 1 km=0.001 Mm

Hence:

510,225,121.4 km^{2} \frac{(0.001 Mm)^{2}}{1 km^{2}}=510.225 Mm^{2}

c) 5.1022(10)^{16} dm^{2}

Knowing 1 Mm=10^{7} dm:

510.225 Mm^{2} \frac{(10^{7} dm)^{2}}{1 Mm^{2}}=5.1022(10)^{16} dm^{2}

7 0
3 years ago
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