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2 years ago

When I say that in Australia there is less rain and is cooler in average, am I talking about weather or climate?

Physics

1 answer:

svetoff [14.1K]2 years ago

8 0

Answer: climate

Explanation:

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RACTIC PTUDIES

Amanda [17]

Answer:

The illumination on the book before the lamp is moved is 9 times the illumination after the lamp is moved.

Explanation:

The distance of the book before the lamp is moved, $d_{b} = 30 cm$

The distance of the book after the lamp is moved, $d_{a} = 90 cm$

Illumination can be given by the formula, $E = \frac{P}{4 \pi d^{2} }$

Illumination before the lamp is moved, $E_{b} = \frac{P}{4 \pi d_{b} ^{2} }$

Illumination after the lamp is moved, $E_{a} = \frac{P}{4 \pi d_{a} ^{2} }$

$\frac{E_{a}}{E_{b}} } = \frac{\frac{P}{4 \pi d_{a} ^{2} } }{\frac{P}{4 \pi d_{b} ^{2} } }$

$\frac{E_{a} }{E_{b} } = \frac{d_{b} ^{2} }{d_{a} ^{2}} \\\frac{E_{a} }{E_{b} } = \frac{30^{2} }{90 ^{2}}\\\frac{E_{a} }{E_{b} } =\frac{900 }{8100}\\\frac{E_{a} }{E_{b} } =\frac{1 }{9}$

$E_{b} = 9E_{a}$

The illumination on the book before the lamp is moved is 9 times the illumination after the lamp is moved.

6 0

3 years ago

Pretend you're

vredina [299]

Answer

What type of phases?

4 0

3 years ago

Is 40,000 °C hotter or colder than 3,600 °C????

Amiraneli [1.4K]

Yes because if 0*c equals 32*F than the higher the number the hotter it is

4 0

3 years ago

Read 2 more answers

Roseanne heated a solution in a beaker as part of a laboratory experiment on energy transfer. After a while, she noticed the liq

sveta [45]

I think it is convection hope I could help.

6 0

3 years ago

What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.

Studentka2010 [4]

Answer: The change in entropy of the given process is 1324.8 J/K

Explanation:

The processes involved in the given problem are:

$1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)$