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3 years ago

7

Explain with one example that different types of organisums live in one habitation

1 answer:

inn [45]3 years ago

5 0

Answer:

When you look at a simple koi pond you can find Koi (the secondary consumer) that feeds off of the zooplankton (first consumer), they eat the phytoplankton (producers). All in a simple food chain

Explanation:

Basically, Koi eat the little animal plankton (zooplankton) that then eats the plant plankton (phytoplankton) that can only end when a part of that habitat is removed. If you got rid of the plant plankton then the whole chain would collapse and most likely die.

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An infinite plane of charge has surface charge density 7.2 Î¼C/m^2. How far apart are the equipotential surfaces whose potential

Rina8888 [55]

Answer:

so the distance between two points are

$d = 0.246 \times 10^{-3} m$

Explanation:

Surface charge density of the charged plane is given as

$\sigma = 7.2 \mu C/m^2$

now we have electric field due to charged planed is given as

$E = \frac{\sigma}{2\epsilon_0}$

now we have

$E = \frac{7.2 \times 10^{-6}}{2(8.85 \times 10^{-12})}$

$E = 4.07 \times 10^5 N/C$

now for the potential difference of 100 Volts we can have the relation as

$E.d = \Delta V$

$4.07 \times 10^5 (d) = 100$

$d = \frac{100}{4.07 \times 10^5}$

$d = 0.246 \times 10^{-3} m$

3 0

3 years ago

Name 3 muscles of the leg

xxTIMURxx [149]

Answer:

calf

Explanation:

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4 0

3 years ago

A 2kg watermelon is dropped from a 4m tall roof a) use the appropriate kinematic equations to determine the instantaneous veloci

Ivenika [448]

Answer:

8.85m/s

Explanation:

The potential energy the watermelon held before dropping is Ep=mgh=2*9.8*4=78.4J.

When it strikes the ground, all of its Ep will transfer into Ek, so 1/2*m*v^2=78.4.

We already knew that m=2, so insert that in, we will get the V^2=78.4 m/s, V=8.85 m/s

6 0

3 years ago

What is the best definition of luminous?

vagabundo [1.1K]

Answer:

the state of giving off light or glow.

8 0

3 years ago

Read 2 more answers

A 55 meter rope is lying on the floor and has has a weight of 4040 N in total. How much work is required to lift up one end of t

Kazeer [188]

Answer:

Explanation:

Given

Length of rope $L=55\ m$

Weight of rope $W=4040\ N$

weight density $\lambda =\frac{4040}{55}=73.45\ N/m$

Work done to lift rope 33 m

$W=\int_{0}^{33}\lambda hdh$

$W=\int_{0}^{33}73.45hdh$

$W=73.45\left [ \left ( \frac{h^2}{2}\right )\right ]^{33}_0$

$W=39.993\ kJ$

6 0

3 years ago