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Vaselesa [24]
3 years ago
12

The process of cytokinesis is different in plant and animal cells. What structure is needed for a plant cell to go through cytok

inesis?
Physics
1 answer:
lesya [120]3 years ago
5 0

Answer:

Cytokinesis occurs differently in animal and plant cells. In animal cells, the plasma membrane of the parent cell pinches inward along the cell's equator until two daughter cells form. In plant cells, a cell plate forms along the equator of the parent cell.

Explanation:

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A 0.106-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find
FrozenT [24]

Answer:

The time rate of change of flux is 1.34 \times 10^{10} \frac{V}{s}

Explanation:

Given :

Current I = 0.106 A

Area of plate A = 36 \times 10^{-4} m^{2}

Plate separation d = 4 \times 10^{-3} m

(A)

First find the capacitance of capacitor,

   C = \frac{\epsilon _{o} A }{d}

Where \epsilon _{o} = 8.85 \times 10^{-12}

   C = \frac{8.85 \times 10^{-12 } \times 36 \times 10^{-4}  }{4 \times 10^{-3} }

   C = 7.9 \times 10^{-12} F

But   C = \frac{Q}{V}

Where Q = It

  C = \frac{It}{V}

  V = \frac{It}{C}

Now differentiate above equation wrt. time,

  \frac{dV}{dt} = \frac{I}{C}

       = \frac{0.106}{7.9 \times 10^{-12} }

       = 1.34 \times 10^{10} \frac{V}{s}

Therefore, the time rate of change of flux is 1.34 \times 10^{10} \frac{V}{s}

8 0
3 years ago
An object thrown vertically upward from the surface of a celestial body at a velocity of 36 ​m/s reaches a height of sequalsminu
Anarel [89]

Answer:

v = -1.8t+36

20 seconds

360 m

40 seconds

36 m/s

The object speed will increase when it is coming down from its highest height.

Explanation:

s=-0.9t^2+36t

Differentiating with respect to time we get

\frac{ds}{dt}=-1.8t+36\\\Rightarrow v=-1.8t+36

a) Velocity of the object after t seconds is v = -1.8t+36

At the highest point v will be 0

0=-1.8t+36\\\Rightarrow t=\frac{-36}{-1.8}\\\Rightarrow t=20\ s

b) The object will reach the highest point after 20 seconds

s=-0.9t^2+36t\\\Rightarrow s=-0.9\times 20^2+36\times 20\\\Rightarrow s=360\ m

c) Highest point the object will reach is 360 m

s=-0.9t^2+36t\\\Rightarrow 360=-0.9t^2+36t\\\Rightarrow -0.9t^2+36t-360=0\\\Rightarrow -9t^2+360t-3600=0

\frac{-360\pm \sqrt{0}}{2\left(-9\right)}\\\Rightarrow t=20\ s

d) Time taken to strike the ground would be 20+20 = 40 seconds

[tex]v=u+at\\\Rightarrow v=0+0.9\times 2\times 20\\\Rightarrow v=36\ m/s

Acceleration will be taken as positive because the object is going down. Hence, the sign changes. 2 is multiplied because the expression is given in the form of s=ut+\frac{1}{2}at^2

e) The velocity with which the object strikes the ground will be 36 m/s

f) The speed will increase when the object has gone up and for 20 seconds and falls down for 20 seconds. The object speed will increase when it is coming down from its highest height.

4 0
3 years ago
A projectile is fired with a velocity of 45 m/s at an angle of 32°. What is the
Dennis_Churaev [7]

Answer:

C

Explanation:

To calculate adjacent of triangle use

\cos(32)  =  \frac{a}{45}

where 45 is the hypotenuse, and a is the adjacent side (horizontal component)

45 \cos(32)  = 38.16

then round to 38.2

8 0
3 years ago
7. A 1300 kg car starts at rest and rolls down a hill from a height of 10.0 m. It then moves across a level (frictionless) surfa
makkiz [27]

Answer:

0.505 m

Explanation:

From the question,

The kinetic energy of the car = energy stored in the spring

1/2mv² = 1/2ke²...................... Equation 1

Where m = mass of the car, v = velocity of the car, k = spring constant of the car, e = extension/compression

make e the subject of the equation

e = v√(m/k)............... Equation 2

We can calculate the value of v, by applying,

v² = u²+2gH...................... Equation 3

Where u = initial velocity of the car, H = height of the car, g = acceleration due to gravity.

Given: u = 0 m/s (from rest), H, 10 m, g = 9.8 m/s²

Substitute into equation 2

v² = 2(10×9.8)

v² = 196

v = √196

v = 14 m/s

Also given: m = 1300 kg, e = 1.0×10⁶ N/m =1000000 N/m

Substitute into equation 2

e = 14√(1300/1000000)

e = 14√(0.00013)

e = 14(0.036)

e = 0.505 m

Hence the maximum distance of the spring is compressed = 0.505 m

5 0
3 years ago
Sunspots. Which of the following is a true statement.
AlladinOne [14]
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6 0
3 years ago
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