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2 years ago

Two internal factors that are important in advancing the cell cycle are

Physics

1 answer:

MA_775_DIABLO [31]2 years ago

3 0

Answer:

Kineses and Cyclin

Explanation:

Kineses and Cyclin are the two important internal factors in the advancing of cell cycle.

In response to a stimuli, the nondirectional motion of an organism or cell depends on the strength of the stimulus or the movement without an apparent physical cause is called Kineses.

Any of a variety of cell division-related proteins that are believed to activate some mitosis processes is called Cyclin.

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A 60 kg student is standing in the train station next to her 10 kg suitcase when her train is called. (A) Estimate how much work

ASHA 777 [7]

Answer

given,

mass of student = 60 Kg

mass of suitcase = 10 Kg

a) Work done by picking of Suitcase is equal to zero

b) acceleration = 0.1 m/s²

distance = 2 m

Using second law conservation

F = m a

F = 10 x 0.1 = 1 N

c) Work done

W = F x s

W = 1 x 2 = 2 J

d) When the are moving with constant speed acceleration is equal to zero

F = m a

F = 10 x 0 = 0 N

W = F x s = 0 x s = 0 J

e) work done = change in kinetic energy

K.E = 2 J

5 0

3 years ago

How do fission nuclear reactions differ from fusion nuclear reactions? Fission reactions involve the conversion of matter into e

MakcuM [25]

How do fission nuclear reactions differ from fusion nuclear reactions?

A. Fission reactions involve the conversion of matter into energy, but fusion reactions do not.

B. Fusion reactions involve the conversion of matter into energy, but fission reactions do not.

C. Fission reactions are used to generate electricity for consumers, but fusion reactions are not.

D. Fusion reactions are used to generate electricity for consumers, but fission reactions are not.

Answer:

Explanation:

Both fission and fusion are nuclear reactions that produce energy, but their applications differs.

Fission is the splitting of a large (heavy, unstable) nucleus into smaller ones, and fusion is the process where nuclei of small atoms are combine together to form the nuclei of larger atoms releasing vast amounts of energy.

The correct answer is c. Fission reactions are used to generate electricity for consumers, but fusion reactions are not.

The physics of fusion is the process that makes the sun shine, and that makes the hydrogen bomb explode.

5 0

3 years ago

Read 2 more answers

The overall length of a piccolo is 32.0 cm. The resonating air column vibrates as in a pipe that is open at both ends. (a) Find

lana66690 [7]

Answer:

lowest frequency = 535.93 Hz

distance between adjacent anti nodes is 4.25 cm

Explanation:

given data

length L = 32 cm = 0.32 m

to find out

frequency and distance between adjacent anti nodes

solution

we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s

lowest frequency will be = $\frac{v}{2L}$ ..............1

put here value in equation 1

lowest frequency will be = $\frac{343}{2(0.32)}$

lowest frequency = 535.93 Hz

and

we have given highest frequency f = 4000Hz

wavelength = $\frac{v}{f}$ ..............2

put here value

wavelength = $\frac{343}{4000}$

wavelength = 0.08575 m

so distance = $\frac{wavelength}{2}$ ..............3

distance = $\frac{0.08575}{2}$

distance = 0.0425 m

so distance between adjacent anti nodes is 4.25 cm

3 0

3 years ago

Suppose you push a hockey puck of mass m across frictionless ice for a time \Delta t, starting from rest, giving thepuck speed v

bazaltina [42]

Answer:

1. $t_2 = 2t_1$

2. $t_2 = t_1\sqrt{2}$

Explanation:

1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.

In Newton's 2nd law: F = ma

where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.

$a_1 = 2a_2$

Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time

$t = v / a_1$

Now that acceleration is halved:

$t = \frac{v}{2a_2}$

$\frac{v}{a_2} = 2t$

You would need to push for twice amount of time $t_2 = 2t_1$

2. The distance traveled by the puck is as the following equation:

$d = at^2$

So if the acceleration is halved while maintaining the same d:

$\frac{d_1}{d_2} = \frac{a_1t_1^2/2}{a_2t_2^2/2}$

As $d_1 = d_2$ , then $d_1/d_2 = 1$ . Also $a_1 = 2a_2$

$1 = \frac{2a_2t_1^2}{a_2t_2^2}$

$t_2^2 = 2t_1^2$

$t_2 = t_1\sqrt{2}\approx 1.14t_1$

So t increased by 1.14

7 0

3 years ago

Ground reaction force acting on carter

mezya [45]

We don't know Carter, and we don't know where he is or what
he's doing, so I'm taking a big chance speculating on an answer.

I'm going to say that if Carter is pretty much just standing there,
or, let's say, lying on the ground taking a nap, then the force of
the ground acting on him is precisely exactly equal to his weight.

8 0

2 years ago