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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

1 answer:

docker41 [41]2 years ago

6 0

Answer:

200 kilocalories

Explanation:

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A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

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3 years ago

Find the work w1 done on the block by the force of magnitude f1 = 95.0 n as the block moves from xi = -5.00 cm to xf = 4.00 cm .

Vlad [161]

<h3><u>Answer;</u></h3>

= 8.55 Joules

<h3><u>Explanation;</u></h3>

Work done is the product of force and the distance moved by an object.

Work done = Force × distance

Force = 95 Newtons

Distance = X2 -X1

= 4 - (-5)

= 9 cm

Thus;

work done = 95 × 9/100

<u>= 8.55 Joules </u>

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Answer:

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