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jeka57 [31]
2 years ago
12

What is most likely the amount of energy available at a trophic level of primary consumers if the amount of energy available to

secondary consumers in that food web is 200 kilocalories?
0 kilocalories
20 kilocalories
200 kilocalories
2,000 kilocalories
Physics
1 answer:
docker41 [41]2 years ago
6 0

Answer:

200 kilocalories

Explanation:

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Give me as many as possible benefits for the neck stretch.
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If you have a neck cramp, so you don't hurt it when working/playing, to loosen the neck muscles
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3 years ago
1. An airow is shot horizontally from a crossbow 1.5 m above the ground. The initial velocity is 45
Amanda [17]

Answer:

Time taken by the arrow to travel along to hit the ground is 0.55 seconds.

Explanation:

The only "force" acting on the "crossbow" to cause it to "hit" the ground is "gravity". There is no initial velocity downward when it shoot.

d=v_{i} t+\frac{1}{2} t^{2}

d = the displacement of the object  

t = the time for which the object moved  

a = acceleration of the object  

v_i = the initial velocity of the object

Given values

d = 1.5 m

t = unknown

a=g=9.8 \mathrm{m} / \mathrm{s}^{2}

\mathrm{V}_{\mathrm{i}}=0 \mathrm{m} / \mathrm{s}

1.5 \mathrm{m}=0(\mathrm{t})+\frac{1}{2}\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right) \mathrm{t}^{2}

1.5=0+4.9 \mathrm{t}^{2}

\mathrm{t}^{2}=\frac{1.5}{4.9}

t^{2}=0.306 \mathrm{s}

Square root both sides

t=\sqrt{0.306}

t = 0.55 s

4 0
3 years ago
If Matthew was traveling into space from Earth, which of these would he be able to reach first? A) Sun B) Venus C) Alpha centaur
const2013 [10]

C.) Alpha Centauri

Explanation:

Due to it being the closest planetary system to earth.

About 4.367 light years away.

3 0
3 years ago
Prove that..<br>please help<br>​
GaryK [48]

\large{ \boxed{ \bf{ \color{red}{Universal \: law \: of \: gravitation}}}}

Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The forces along the line joining the centre of the two objects.

❍ Let us consider two masses m1 and m2 line at a separation distance d. Let the force of attraction between the two objects be F.

According to universal law of gravitation,

\large{ \longrightarrow{ \rm{F \propto m_1 m_2}}}

Also,

\large{ \longrightarrow{ \rm{ F \propto  \dfrac{1}{ {d}^{2} } }}}

Combining both, We will get

\large{ \longrightarrow{ \rm{F  \propto  \dfrac{ m_1 m_2}{ {d}^{2}}}}}

Or, We can write it as,

\large{ \longrightarrow{ \rm{F  \propto  \:  G \dfrac{ m_1 m_2}{ {d}^{2} }}}}

Where, G is the constant of proportionality and it is called 'Universal Gravitational constant'.

☯️ Hence, derived !!

<u>━━━━━━━━━━━━━━━━━━━━</u>

8 0
3 years ago
How might you describe the mathematical procedure of finding the displacement when an object travels in two opposite directions?
levacccp [35]
Displacement is a vector quantity. So, you incorporate the vector calculations when you try to determine the resultant vector. This is the shortest path from the starting point to the endpoint. If they are moving on one axis only, you use sign conventions. For motions moving to the left, use the negative sign. If it's moving to the right, then use the positive sign. Now, it the object moves 2 km to the left, and 2 km also to the right, the displacement is zero.

Displacement = 2 km - 2km = 0

Generally, the equation is:
<span>Displacement = Distance of motion to the right - Distance of motion to the left</span>
4 0
3 years ago
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