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3 years ago

Energy that is passed on from one trophic level to the next is called what?

1 answer:

8 0

Energy decreases as it moves uptrophic levels because energy is lost as metabolic heat when the organisms from one trophic level are consumed by organisms from the next level.Trophic level transfer efficiency (TLTE) measures the amount of energy that is transferred between trophic levels.

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What is the answer I’m stuck on this one it’s a screenshot for the question

Dmitry_Shevchenko [17]

Answer:

deceleration is the opposite of acceleration

Explanation:

We know that acceleration is the increase of speed with respect to time. So deceleration must be represented on the graph as a decrease in speed over time.a

3 0

2 years ago

(NEED HELP PLEASE) A physics student goes to the roof of the school, 24.15 m above the ground, and drops a pumpkin straight down

slavikrds [6]

Answer:

t = 2.2 s

Explanation:

Given that,

Height of the roof, h = 24.15 m

The initial velocity of the pumpkin, u = 0

We need to find the time taken for the pumpkin to hit the ground. Let the time be t. Using second equation of kinematics to find it as follows :

$h=ut+\dfrac{1}{2}at^2$

Here, u = 0 and a = g

$h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 24.15}{9.8}} \\\\t=2.22\ s$

So, it will take 2.22 s for the pumpkin to hit the ground.

7 0

3 years ago

A bystander observes the musicians heading toward each other. When musician #1 is 100 m away, the intensity is 1.24 x 10-8 W/m^2

777dan777 [17]

Explanation:

Given that,

Distance 1, r = 100 m

Intensity, $I_1=1.24\times 10^{-8}\ W/m^2$

If distance 2, r' = 25 m

We need to find the intensity and the intensity level at 25 meters. Intensity and a distance r is given by :

$I=\dfrac{P}{4\pi r^2}$ .........(1)

Let I' is the intensity at r'. So,

$I'=\dfrac{P}{4\pi r'^2}$ ............(2)

From equation (1) and (2) :

$I'=\dfrac{Ir}{r'^2}$

$I'=\dfrac{1.24\times 10^{-8}\times 100}{25^2}$

$I'=1.98\times 10^{-9}\ W/m^2$

Intensity level is given by :

$dB=10\ log(\dfrac{I'}{I_o})$ , $I_o=10^{-12}\ W/m^2$

$dB=10\ log(\dfrac{1.98\times 10^{-9}}{10^{-12}})$

dB = 32.96 dB

Hence, this is the required solution.

7 0

3 years ago

A rock is thrown horizontally from a building at 15 m/s. It hits the ground 45 m from the base of the building. How high was the

evablogger [386]

Answer:

Explanation:

hope this helps!

7 0

3 years ago

A sample of metallic frewium weighs 185N on a spring scale in air. When immersed in pure water, the frewium pulls on the scale w

balu736 [363]

Wow ! This one could have some twists and turns in it.
Fasten your seat belt. It's going to be a boompy ride.

-- The buoyant force is precisely the missing 30N .

-- In order to calculate the density of the frewium sample, we need to know
its mass and its volume. Then, density = mass/volume .

-- From the weight of the sample in air, we can closely calculate its mass.

 Weight = (mass) x (gravity)
 185N = (mass) x (9.81 m/s²)
 Mass = (185N) / (9.81 m/s²) = 18.858 kilograms of frewium

-- For its volume, we need to calculate the volume of the displaced water.

The buoyant force is equal to the weight of displaced water, and the
density of water is about 1 gram per cm³. So the volume of the
displaced water (in cm³) is the same as the number of grams in it.

The weight of the displaced water is 30N, and weight = (mass) (gravity).

 30N = (mass of the displaced water) x (9.81 m/s²)

 Mass = (30N) / (9.81 m/s²) = 3.058 kilograms

 Volume of displaced water = 3,058 cm³

Finally, density of the frewium sample = (mass)/(volume)

 Density = (18,858 grams) / (3,058 cm³) = 6.167 gm/cm³ (rounded)

================================================

I'm thinking that this must be the hard way to do it,
because I noticed that

 (weight in air) / (buoyant force) = 185N / 30N = 6.1666...

So apparently . . .

 (density of a sample) / (density of water) =

 (weight of the sample in air) / (buoyant force in water) .

I never knew that, but it's a good factoid to keep in my tool-box.

3 0

3 years ago