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kakasveta [241]
3 years ago
12

Which two structures are not found in animal cells?

Physics
1 answer:
Alinara [238K]3 years ago
7 0

Answer:

C

Explanation:

So, lets start off by thinking about what we know is in both cells, and rule out from there.

There are just some basic things you need in eukaryotic cells, those include the following:

  • Nucleus
  • cell membrane
  • mitochondria
  • cytoplasm(dont get confused with chloroplast)
  • Rough and smooth endoplasmic reticulum
  • Golgi vescile
  • Golci apparatus
  • Nucleolus(Dont get confused with nulceus)

Here are jsut a few examples of what can be found in both plant and animal cells, however we can already rule out all of the following except for:

C

Because remember, c talks about the chlorplast, which is not the chlorplasm found in both plant anf animal cells, its just the chloroplast found in plant cells.

Although I gave the asnwer to you. Here are some things that are in plant cells that arent in animal cells:

  • Cell wall
  • vacioles
  • central vaciole
  • chlroplast
  • plastids

Animal cells on the other hand, have a few things that plant cells do not as well:

  • Lysome
  • Centrolsome

Hope this helps! ;)

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Four weightlifters (A-D) enter a competition. The mass, distance, and time of their lifts are shown in the table.
siniylev [52]

Let Pa, Pb, Pc, and Pd be the powers delivered by weightlifters A, B, C, and D, respectively.

Use this equation to determine each power value:

P = W÷Δt

P is the power, W is the work done by the weightlifter, and Δt is the elapsed time.

A) Determining Pa:

Pa = W÷Δt

The weightlifter does work to lift the weight up a certain distance. Therefore the work done is equal to the weight's gain in gravitational potential energy. The equation for gravitational PE is

PE = mgh

PE is the potential energy, m is the mass of the weight, g is the acceleration of objects due to earth's gravity, and h is the distance the weight was lifted.

We can equate W = PE = mgh, therefore we can make the following substitution:

Pa = mgh÷Δt

Given values:

m = 100.0kg

g = 9.81m/s²

h = 2.25m

Δt = 0.151s

Plug in the values and solve for Pa

Pa = 100.0×9.81×2.25÷0.151

<u>Pa = 14600W</u> (watt is the SI derived unit of power)

B) Determining Pb:

Let us use our new equation derived in part A to solve for Pb:

Pb = mgh÷Δt

Given values:

m = 150.0kg

g = 9.81m/s²

h = 1.76m

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<u>Pb = 49800W</u>

C) Determining Pc:

Pc = mgh÷Δt

Given values:

m = 200.0kg

g = 9.81m/s²

h = 1.50m

Δt = 0.217s

Plug in the values and solve for Pc

Pc = 200.0×9.81×1.50÷0.217

<u>Pc = 13600W</u>

D) Determining Pd:

Pd = mgh÷Δt

Given values:

m = 250.0kg

g = 9.81m/s²

h = 1.25m

Δt = 0.206s

Plug in the values and solve for Pd

Pd = 250.0×9.81×1.25÷0.206

<u>Pd = 14900W</u>

Compare the following power values:

Pa = 14600W, Pb = 49800W, Pc = 13600W, Pd = 14900W

Pc is the lowest value.

Therefore, weightlifter C delivers the least power.

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