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The amount of energy lost at the transition between each trophic level of the pyramid of energy is about _______. a. 10% b. 50%

Sophie [7]

The amount of energy lost at the transition between each trophic level of the pyramid of energy is about 90%.

There are a total of 4 trophic levels.

The producers which bare plants represent the first trophic level. Herbivores represent the second trophic level. Carnivores represent the third trophic level. Top carnivores represent the fourth trophic level.

The 10 % rule is followed by the energy flow in a food chain. It means that moving from one trophic level to another, only 10% of energy is transferred and the rest is lost in the atmosphere.

So the 90% percent energy is lost at the transition between each trophic level of the pyramid of energy.

If you need to learn more about the trophic levels, click here

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3 0

1 year ago

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A particle with a charge of 2e moves between two points which have a potential difference of 75V. What is the change in potentia

Sonbull [250]

Electric potential energy is defined as Ep=Q*V where Q is the magnitude of the charge and V is the potential difference. So when a charge moves between the points that have a potential difference, it's energy changes.

In our case:

Q=2e=2*(-1.6*10^-19) C
V=75 V

Ep=(-3.2*10^-19)*75

Ep=-2.4*10^-17 J

The change in potential energy of the charge is -2.4*10^-17 J

5 0

3 years ago

Two workers pull horizontally on a heavy box. but one pulls twice as hard as the other. The larger pull is directed at 21.0° wes

pantera1 [17]

Answer:

The magnitude of F1 is

$|F1|=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

Explanation:

<u>Net Force </u>

Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.

The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.

Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

$\displaystyle -2F\ sin21^o+F\ cos\alpha =0$

The sum of the vertical components of F1 and F2 must equal the total force Ft

$\displaystyle -2F\ cos21^o+F\ sin\alpha =460$

Solving for $\alpha$ in the first equation

$\displaystyle cos\alpha =\frac{2F\ sin21^o}{F}=2sin21^o$

$\displaystyle cos\alpha =0.717=>\alpha =44.214^o$

$\displaystyle F(2cos21^o+sin\alpha)=460$

$\displaystyle F=\frac{460}{2cos21^o+sin\alpha}$

$\displaystyle F=\frac{460}{2cos21^o+sin44.214^o}$

$\displaystyle F=179.37\ N$

The magnitude of F1 is

$|F1|=2*F=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

4 0

3 years ago

What is any factor of an experiment that is not changed called? control, dependent variable ,independent variable

Softa [21]

A control is the factor which is not changed

5 0

3 years ago

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An 800-g block of ice at 0.00°C is resting in a large bath of water at 0.00°C insulated from the environment. After an entropy c

Allisa [31]

Answer:

Unmeltedd ice = 308.109 g

Explanation:

Gibbs Free energy:

A systems Gibbs Free Energy is defined as the free energy of the product of the absolute temperature and the entropy change less than the enthalpy change.

Therefore, G = ΔH-TΔS

where G is Gibbs Free Energy

ΔH is enthalpy change

T is absolute temperature

ΔS is entropy change

Here since there is a phase change, therefore G will be 0.

∴ΔH = TΔS

Given: Temperature, T = 0°C = 273 K

Entropy change,ΔS = 600 J/K

Latent heat of fusion of water = 333 J/g

∴ΔH = TΔS

∴ΔH = 273 x 600

= 163800 J

So this is the amount of enthalpy that will be used into melting of ice.

∴ΔH = mass of ice melted x latent heat of fusion of water

Mass of ice melted = ΔH / latent heat of fusion of water

= 163800 / 333

= 491.891 g

This is the mass of ice melted.

And initial amount of ice is 800 g

Amount of ice left after melting = Initial amount of ice - amount of ice melted

= 800-491.891

= 308.109 g

Amount of ice remained after melting = 308.109 g

8 0

3 years ago