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andre [41]
3 years ago
11

Name four biotic factors

Physics
2 answers:
DiKsa [7]3 years ago
5 0
Biotic factors include animals, plants, fungi, bacteria, and protists. Some examples of abiotic factors are water, soil, air, sunlight, temperature, and minerals.
jonny [76]3 years ago
3 0

Explanation:

(1) predation

(2) competition

(3) autotrophs

(4) heterotrophs

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How much heat energy (in megajoules) is needed to convert 7 kilograms of ice at –9°C to water at 0°C?
Hitman42 [59]
The heat  required to change the phase of a substance can be calculated by using the formula,

q = mCΔT where q is the heat needed, m is the mass of the substance, C is the specific heat capacity and ΔT is the change in temperature.

q = 7000 g (4.18 J/ g °C) (0°C - (-9°C))
q = 263340 J or .26334 MJ
Hope it's correct! ( If so rank as brainliest answer) :)
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4 years ago
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Ymorist [56]

Answer:

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Explanation: ;0

6 0
3 years ago
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Hi, I am having some difficulties in solving this question. Could someone please explain this question to me in detail. The thin
aniked [119]

Answer:

a) 3.0×10⁸ m

b) 0 m

Explanation:

Displacement is the distance from the starting position to the final position.

a) In half a year, the Earth travels from one point on the circle to the point on the exact opposite side of the circle (from 0° to 180°).  The distance between the points is the diameter of the circle.

x = 2r

x = 2 (1.5×10⁸ m)

x = 3.0×10⁸ m

b) In a full year, the Earth travels one full revolution, so it ends up back where it started.  The displacement is therefore 0 m.

8 0
4 years ago
Which of the following is the flow of electrons through a wire or a conductor?
Sonja [21]

Explanation:

electric current is the answer

7 0
3 years ago
The average distance of the planet mercury from the sun is 0.39 times the average distance of the earth from the sun. How long i
Stels [109]

Answer:

T_1=0.24y

Explanation:

Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:

(\frac{T_1}{T_2})^2=(\frac{D_1}{D_2})^3

Where T_1 and T_2 are the orbital periods of Mercury and Earth respectively. We have D_1=0.39D_2 and T_2=1y. Replacing this and solving for

T_1^2=T_2^2(\frac{D_1}{D_2})^3\\T_1^2=(1y)^2(\frac{0.39D_2}{D_2})^3\\T_1^2=1y^2(0.39)^3\\T_1^2=0.059319y^2\\T_1=0.24y

5 0
3 years ago
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