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Westkost [7]
3 years ago
5

Help me plzzz?? I need help it on the picture

Physics
1 answer:
madam [21]3 years ago
3 0

No. 
The acceleration of gravity on or near Earth's surface is 9.8 m/s² ,
not 20 m/s² .

If it were 20 m/s², then you would weigh almost exactly double
what you really weigh now.
 
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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

brainly.com/question/13267084

#SPJ4

3 0
1 year ago
Read 2 more answers
When a fixed amount of ideal gas goes through an isobaric expansion A) its internal (thermal) energy does not change.B) the gas
Bingel [31]
<h2>Answer: its temperature must increase.</h2>

Explanation:

In an isobaric process the pressure remains constant, which means the initial pressure and the final pressure will be the same.

In addition, during this thermodynamic process, the volume of the ideal gas expands or contracts in such a way that the variation of pressure \Delta P is neutralized.

Now, according to the First law of Thermodynamics that establishes the conservation of energy:

\Delta U=\Delta Q-\Delta W   (1)

Where:

\Delta U is the internal energy

\Delta Q is the heat transferred

\Delta W is the work

Now, for an isobaric process:

\Delta W=P\Delta V    (2)

Where:

P is the pressure (<u>always positive</u>)

\Delta V is the volume variation of the gas

<u />

<u>Here we have two possible results:</u>

-If the gas expands (positive \Delta V), the work is positive.

-If the gas compresses (negative \Delta V), the work is negative.

In this case we are talking about the first result (work is positive).

Then, according to the above, equation (1) can be written as follows:

\Delta U=\Delta Q - P\Delta V   (3)

Clearing \Delta Q:

\Delta Q=\Delta U+P \Delta V    (4)

Then, for an ideal gas in an isobaric process, part of the heat (Q) added to the system will be used to do work (positive in this case) and the other part <u>will increase the internal energy</u>, hence <u>the temperature will increase as well.</u>

7 0
3 years ago
Which biome has the most variable year round temperature
skad [1K]
That should be the Grasslands ?
5 0
2 years ago
Problem:
gayaneshka [121]
This is confusing yes
3 0
3 years ago
Two metra trains approach each other on separate but parallel tracks. one has a speed of 90 km/hr, the other 80 km/hr. initially
Gennadij [26K]

The trains take <u>57.4 s</u> to pass each other.

Two trains A and B move towards each other. Let A move along the positive x axis and B along the negative x axis.

therefore,

v_A=90 km/h\\ v_B=-80 km/h

The relative velocity of the train A with respect to B is given by,

v_A_B=v_A-v_B\\ =(90km/h)-(-80km/h)\\ =170km/h

If the train B is assumed to be at rest, the train A would appear to move towards it with a speed of 170 km/h.

The trains are a distance d = 2.71 km apart.

Since speed is the distance traveled per unit time, the time taken by the trains to cross each other is given by,

t= \frac{d}{v_A_B}

Substitute 2.71 km for d and 170 km/h for v_A_B

t= \frac{d}{v_A_B}\\ =\frac{2.71 km}{170 km/h} \\ =0.01594 h

Express the time in seconds.

t=(0.01594h)(3600s/h)=57.39s

Thus, the trains cross each other in <u>57.4 s</u>.

6 0
3 years ago
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