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dalvyx [7]
3 years ago
10

can cuclin concentration durng mitosis be explained by the fact that the cell divides in two and thus divides the material in th

e cell into two smaller portions?
Physics
1 answer:
lana [24]3 years ago
5 0

Answer: No

Can the change in cyclin concentration during mitosis be explained by the fact that the cell divides in two and thus divides the material in the cell into two smaller volumes?

Explanation: The cyclin concentration is not halved but degraded during mitosis.

There is an increase in cyclin concentrations at interphase. These changes are caused by the presence of Cyclin Dependent Kinase (CDK) complexes. CDK being a substrate of cyclin catalyses cyclin, thereby increasing its concentration. During mitosis, cyclins are destroyed, signifying the end of mitosis and cytokinesis.

Without this process, it will be impossible for the cell to exit mitosis.

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Thorium-232 decays by alpha decay. Part of the nuclear equation is shown below. Fill in the blank with a number. 232/90
UkoKoshka [18]

its 228 on my apex......

8 0
3 years ago
The combining of light nuclei is called blank. Blank as in ____. Not actually blank. You know what im tryin to say.
Andreas93 [3]

Answer:

The combining of light nuclei is called nuclear fusion.

6 0
3 years ago
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A 10.0-kg box starts at rest on a level floor. An external, horizontal force of 2.00 × 102 N is applied to the box for a distanc
Harman [31]

Answer:

vf = 11.2 m/s

Explanation:

m = 10 Kg

F = 2*10² N

x = 4.00 m

μ = 0.44

vi = 0 m/s

vf = ?

We can apply Newton's 2nd Law

∑ Fx = m*a   (→)

F - Ffriction = m*a  ⇒  F - (μ*N) = F - (μ*m*g) = m*a   ⇒  a = (F - μ*m*g)/m

⇒    a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²

then , we use the equation

vf² = vi² + 2*a*x    ⇒    vf = √(vi² + 2*a*x)

⇒   vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s

7 0
3 years ago
A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040
Minchanka [31]

Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force (F_{B}), as follows:

F_{c} = F_{B} \rightarrow \frac{m*v^{2}}{r} = qvB (1)

<u>Where:</u>

<em>m: is the proton's mass =  1.67*10⁻²⁷ kg</em>

<em>v: is the proton's velocity</em>

<em>r: is the radius of the proton's orbit</em>

<em>q: is the proton charge = 1.6*10⁻¹⁹ C</em>

<em>B: is the magnetic field = 0.040 T </em>

Solving equation (1) for r, we have:

r = \frac{mv}{qB}   (2)

By conservation of energy, we can find the velocity of the proton:

K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V   (3)

<u>Where:</u>

<em>K: is kinetic energy</em>

<em>U: is electrostatic potential energy</em>

<em>ΔV: is the potential difference = 1.0 kV </em>

Solving equation (3) for v, we have:

v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s  

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!  

5 0
3 years ago
Which of the following is not a form of an inclined plane?
disa [49]

Answer: piece of paper cut into a triangle  

Explanation:

A plane surface lifted at an angle to horizontal from one edge is said to be an inclined plane. A wedge, a ramp or a screw are examples of a forms of inclined plane. A screw is like a inclined plane wound on a cylinder. These are simple machines. On the other hand, a piece of paper cut into a triangle is not an inclined plane as it does rise at an angle from horizontal. It is not a machine.

8 0
3 years ago
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