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Hatshy [7]
3 years ago
10

D) cube

Physics
2 answers:
kondor19780726 [428]3 years ago
8 0

Answer:

c because it makes since

Explanation:

I took the test

Zielflug [23.3K]3 years ago
5 0
C)24cm
Explanation: I took the test!
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<span>v(4 seconds)= 300 m/s - 9.8 (m/s^2)(4s) = 260.8 m/s </span>, hope this helps:)
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Which trophic level has the least available energy in kilojoules in this food web?
insens350 [35]

The highest trophic level has the least available energy in kilojoules.

Even though the food web is not shown in the question, but we know that energy decreases steadily as it is passed on from one trophic level to the next according to the second law of thermodynamics.

Energy enters into the system from the sun. The primary producers utilize this energy to produce food. As plants are eaten by animals, this energy is transferred along the food web an diminishes at each higher trophic level.

At the highest trophic level, the the least available energy in kilojoules in this food web is found.

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murzikaleks [220]

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7 0
3 years ago
The tires on your truck have 0.35 m radius. In a straight line, you drive 2600 m. What is the angular displacement of the tire,
Travka [436]
1) In a circular motion, the angular displacement \theta is given by
\theta =  \frac{S}{r}
where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire: S=2600 m. Using the information about the radius, r=0.35 m, we find the total angular displacement:
\theta =  \frac{2600 m}{0.35 m} =7428 rad

2) If we put larger tires, with radius r=0.60 m, the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
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4 years ago
Gauss’ law: a. Relates the surface charge density to the electric field.b. Relates the electric field at points on a closed surf
Mamont248 [21]

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b. Relates the electric field at points on a closed surface to the net charge enclosed by that surface

Explanation:

Gauss's law states that the flux of certain fields through a closed surface is proportional to the magnitude of the sources of that field within the same surface. The electric flux expresses the measure of the electric field that crosses a certain surface. Therefore, the electric field on a closed surface is proportional to the net charge enclosed by that surface.

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