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Korolek [52]
3 years ago
14

Biotic parts of an ecosystem interact with

Physics
1 answer:
natta225 [31]3 years ago
6 0
Biotic parts of an ecosystem interact with living things like plants, animals, bacteria, and more organisms. Abiotic parts of an ecosystem are non-living things 
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It’s because flourecent lights operate at higher temperatures than incadecent lights.

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Is friction and pushing similar ????
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Yes, they are.

Explanation:

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A coffee filter of mass 1.2 grams dropped from a height of 1 m reaches the ground with a speed of 0.8 m/s. How much kinetic ener
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<u>Answer:</u> The energy gained by the air molecules is 0.011 J.

<u>Explanation:</u>

Law of conservation of energy states that energy can neither be created nor be destroyed but it can only be transformed from one form to another form.

Here, the potential energy of the coffee filter is getting converted into kinetic energy of the coffee filter and some energy is lost by it which is gained by the air molecules in the form of kinetic energy.

So, calculating the potential energy of coffee filter, we use the equation:

P = mgh

where,

m = mass of coffee filter = 1.2 g = 1.2\times 10^{-3}kg    (Conversion used: 1 kg = 1000 g)

g = acceleration due to gravity = 9.8m/s^2

h = height of coffee filter = 1 m

Putting values in above equation, we get:

P=1.2\times 10^{-3}kg\times 9.8m/s^2\times 1m\\\\P=1.176\times 10^{-2}J

  • Calculating the kinetic energy of coffee filter, we use the equation:

E=\frac{1}{2}mv^2

where,

m = mass of coffee filter = 1.2 g = 1.2\times 10^{-3}kg

v = speed of coffee filter = 0.8 m/s

Putting values in above equation, we get:

E=\frac{1}{2}\times 1.2\times 10^{-3}kg\times (0.8m/s)^2\\\\E=3.84\times 10^{-4}J

As, energy lost by coffee filter = energy gained by air molecules

So, energy lost by coffee filter = Potential energy - Kinetic energy

Energy lost by coffee filter = (1.176\times 10^{-2})-(3.84\times 10^{-4})=0.011J

Hence, the energy gained by the air molecules is 0.011 J.

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3 years ago
A parallel-plate capacitor is connected to a battery. What happens to the stored energy if the plate separation is doubled while
marta [7]

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(c)  As 'd' becomes doubled, energy decreases by the factor of 2

Explanation:

Energy stored in a parallel plate capacitor is given by:

U=\frac{1}{2}CV^2\\\\C=\frac{A\epsilon_{o}}{d}\\\\then\\\\U=\frac{1}{2}\frac{A\epsilon_{o}}{d}V^2--(1)\\\\

As capacitor remains connected to the battery so V remains constant. As can be seen from (1) that energy is inversely proportional to the separation between the plates so as 'd' becomes doubled, energy decreases by the factor of 2.

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Meiosis is the correct answer because of the fact all the other answers are no where near correct
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