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2 years ago

If carrots have 80j of energy how much energy will a rabbi get from eating the carrot

Physics

1 answer:

beks73 [17]2 years ago

8 0

Answer:

Explanation:

According to the 10% law of energy transfer,

only about 10% of energy will be transferred to

the successive level from the previous trophic

level. Therefore according to this law carrot will

have the highest energy level as it is present at

the lowest trophic level.

Then it will transfer 10% of its consumed energy

to the rabbit and then snakes will get 10%

energy of rabbit after consuming it. Therefore if

the energy amount is given it can be calculated

by dividing it by 10 for upper trophic level.

So if a rabbit receives 80 calories of energy

from eating carrots it will pass only 8

(80/10=8) calory of energy on to a snake.

Therefore the correct answer is 8 calories

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4 years ago

Usain Bolt is the fastest human at 12.4 m/s. If Usain Bolt had a mass of 94kg, what was Usain Bolt’s KE?

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3 years ago

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4 years ago

At locations A and B, the electric potential has the values VA=1.51 VVA=1.51 V and VB=5.81 V,VB=5.81 V, respectively. A proton r

Oksana_A [137]

Answer:

For proton:

A. The proton is released from Vb (highest potential)

B. v = 2.9x10⁴ m/s

For electron:

A. The electron is released from Va (lowest potential)

B. v = 1.2x10⁶ m/s

Explanation:

For a proton we have:

A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.

Having that:

Va = 1.51 V and Vb = 5.81 V

We can see that the proton moves from Vb to Va, hence the proton was released from Vb.

B. We now that the work done by an electric field is given by:

$W = \Delta Vq$ (1)

Where:

q: is the proton's charge = 1.6x10⁻¹⁹ C

V: is the potential

Also, the work is equal to:

$W = \Delta K = (K_{a} - K_{b}) = \frac{1}{2}mv_{a}^{2} - \frac{1}{2}mv_{b}^{2}$ (2)

Where:

K: is the kinetic energy

m: is the proton's mass = 1.67x10⁻²⁷ kg

$v_{a}$ : is the velocity in the point a

$v_{b}$ : is the velocity in the point b = 0 (starts from rest)

Matching equation (1) with (2) we have:

$\Delta Vq = \frac{1}{2}mv_{a}^{2}$

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}1.67 \cdot 10^{-27} kg*v_{a}^{2}$

$v_{a} = 2.9 \cdot 10^{4} m/s$

For an electron we have:

A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.

B. The speed is:

$\Delta Vq = \frac{1}{2}mv_{b}^{2} - \frac{1}{2}mv_{a}^{2}$

Since $v_{a}$ = 0 (starts from rest) and $m_{e}$ = 9.1x10⁻³¹ kg (electron's mass), we have:

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}9.1 \cdot 10^{-31} kg*v_{b}^{2}$

$v_{b} = 1.2 \cdot 10^{6} m/s$

I hope it helps you!

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4 years ago

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torisob [31]

Answer:

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3 years ago