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san4es73 [151]
3 years ago
9

Each level in a food chain contains less energy than the one below it because some energy is

Physics
1 answer:
Ira Lisetskai [31]3 years ago
7 0

Answer:

Because some of the energy is dissipated in the metabolic activities such as digestion and maintaining the warmth of the body of the organism at each level.

Explanation:

Food chain is the sequential chain of energy transfer from one onrgaanism at a trophic level to another organism at the other trophic level. Autotrophs receive the energy from the sun and prepare their own food. Autotrophs then get consumed by the primary heterotrophs and further primary heterotrophs by the secondary heterotrophs. When the secondary heterotrophs die they are then consumed by the scavangers and decayed by the micro-organisms in to the soil which finally help the autotrophs (plants) in their growth.

At each trophic level of the food chain the energy does not gets completely transferred because the organisms also spend some energy in search of the food and the processing of food is also an energy consuming process.

The energy is also used in combating the climatic conditions and predators for the survival.

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Alex787 [66]

Answer: The force constant k is 10600 kg/s^2

Step by step:

Use the law of energy conservation. When the elevator hits the spring, it has a certain kinetic and a potential energy. When the elevator reaches the point of still stand the kinetic and potential energies have been transformed to work performed by the elevator in the form of friction (brake clamp) and loading the spring.

Let us define the vertical height axis as having two points: h=2m at the point of elevator hitting the spring, and h=0m at the point of stopping.

The total energy at the point h=2m is:

E_{tot}=E_{kin}+E_{pot}\\E_{tot}= \frac{1}{2}mv^2+mg\Delta h = \frac{1}{2}2000 kg 4^2\frac{m^2}{s^2}+2000kg\, 9.8\frac{m}{s^2}2m=55200\,kg\frac{m^2}{s^2}

The total energy at the point h=0m is:

E_{tot}=E_{kin}+E_{pot}+Work=0+0+ Work\\E_{tot} =F_{friction}\Delta h+\frac{1}{2}k (\Delta h)^2=17000N\cdot 2m+\frac{1}{2}k\cdot 2^2 m^2

The two Energy values are to be equal (by law of energy conservation), which allows us to determine the only unknown, namely the force constant k:

17000N\cdot 2m+\frac{1}{2}k\cdot 2^2 m^2 = 55200 \,kg\frac{m^2}{s^2}\\k = \frac{55200-34000}{2}\,\frac{kg}{s^2}=10600\frac{kg}{s^2}

5 0
3 years ago
Block A, with a mass of 4.0 kg, is moving with a speed of 2.0 m/s while block B , with a mass of 8.0 kg, is moving in the opposi
Anon25 [30]
Consider velocity to the right as positive.

First mass:
m₁ = 4.0 kg
v₁ = 2.0 m/s to the right

Second mass:
m₂ = 8.0 kg
v₂ = -3.0 m/s to the left

Total momentum of the system is
P = m₁v₁ + m₂v₂
   = 4*2 + 8*(-3)
  = -16 (kg-m)/s

Let v (m/s) be the velocity of the center of mass of the 2-block system.

Because momentum of the system is preserved, therefore
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v = -1.333 m/s

Answer:
The center of mass is moving at 1.33 m/s to the left.
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