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3 years ago

Which stage of the cell cycle involves the division of the cytoplasm?

Physics

2 answers:

sergij07 [2.7K]3 years ago

7 0

Answer:

Explanation:

I just took the question on edge, you can also remember cytokinesis is the division of the cytoplasm by the c at the beginning think of the word cut.

Dmitry_Shevchenko [17]3 years ago

7 0

OK my friend the best answer to this question will be Cytokineses <span />

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Can someone help me?

Alik [6]

Answer:

Resultant = 3.05N

Explanation:

$r = \sqrt{{5}^{2} + 5^{2} - 2(5)(5) \cos(120) }$

$r = \sqrt{25 + 25 - 50(0.8142)}$

$r = \sqrt{50 - 40.71} = \sqrt{9.29}$

$r = 3.05$

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2 years ago

You just explored work and heat. Now test your understanding,

Lyrx [107]

Answer:

the hotter it gets, the liquid(mercury), expands more and more, and will rise up the tube to the correct line to read the tempature

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2 years ago

The largest hailstone every measured fell in Vivian, Nebraska in 2010. The circumference of that hailstone was 19 inches. Using

Crazy boy [7]

Answer:

6.04788 in

$115.82654\ in^3$

78.38779 m/s

0.88159 kg

34.55294 J

Explanation:

Circumference is given by

$c=2\pi r\\\Rightarrow r=\dfrac{c}{2\pi}\\\Rightarrow r=\dfrac{19}{2\pi}\\\Rightarrow r=3.02394\ in$

Diameter is given by

$d=2r\\\Rightarrow d=2\times 3.02394\\\Rightarrow d=6.04788\ in$

The diameter is 6.04788 in

$6.04788\times 2.54=15.3616152\ cm$

Volume of sphere is given by

$v=\dfrac{4}{3}\pi r^3\\\Rightarrow v=\dfrac{4}{3}\pi 3.02394^3\\\Rightarrow v=115.82654\ in^3$

The volume is $115.82654\ in^3$

$115.82654\times \dfrac{1}{1728}=0.06702\ ft^3$

Fall velocity is given by

$V=k\sqrt{d}\\\Rightarrow V=20\sqrt{15.3616152}\\\Rightarrow V=78.38779\ m/s$

The velocity of the fall will be 78.38779 m/s

Mass is given by

$m=\rho v\\\Rightarrow m=29\times 0.06702\\\Rightarrow m=1.94358\ lb$

$1.94358\ lb=1.94358\times \dfrac{1}{2.20462}=0.88159\ kg$

The mass is 0.88159 kg

Kinetic energy is given by

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}0.88159\times 78.38779\\\Rightarrow K=34.55294\ J$

The kinetic energy is 34.55294 J

4 0

2 years ago

Two rock climbers, bill and karen, use safety ropes of similar length. karen's rope is more elastic, called a dynamic rope by cl

Oliga [24]

You never gave us an answer, so therefore, we can't answer this question.

3 0

3 years ago

Read 2 more answers

A Cessna 150 aircraft has a lift-off speed of approximately 125 km/h. What minimum constant acceleration does this require if th

Marysya12 [62]

Answer:

Mininmum acceleration required = $2.74 m/s^{2}$

Explanation:

Given,

Lift off speed = 125km/hr.

We know km/hr conversion factor to m/s is $\frac{5}{18}$ ,

Therefore,

Lift off speed required = $125 \times\frac{5}{18}$ m/s

Lift off speed required = 34.72222 m/s

Length of Takeoff Run given = 220 m.

So,

the flight needs to achieve its takeoff speed at a constant acceleration before it reaches the end of the 220 m runway.

Consider the formula from kinematics,

$v^{2}-u^{2}=2as$ ,

where,

v - final velocity of particle,

u - initial velocity of particle

a - the constant acceleration at which the particle is moving with

s - distance travelled by particle

So at minimum acceleration,it reaches the takeoff speed at the end of the runway.

Therefore in the formula,

we use, v = Lift off speed = 34.72222 m/s;

u = 0 m/s (plane starts from rest);

a = minimum acceleration of the plane

s = 220 m;

Substituting these values in the formula, we get,

$(34.72222)^{2}-0 = 2\times a \times 220$

a = $\frac{(34.72222)^{2} }{2\times220}$

a = $2.74 m/s^{2}$ .

Therefore,the minimum acceleration of the plane required = $2.74 m/s^{2}$ .

7 0

3 years ago