Describe the process of sexual reproduction in terms of the life cycle of cells.

1 answer:

5 0

The first growth phase (G1): During the G1 stage, the cell doubles in size and doubles the number of organelles.

The synthesis phase (S): The DNA is replicated during this phase. In other words, an identical copy of all the cell’s DNA is made. This ensures that each new cell has a set of genetic material identical to that of the parental cell. This process is called DNA replication.

The second growth phase (G2): Proteins are synthesized that will help the cell divide. At the end of interphase, the cell is ready to enter mitosis.

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what is the resistance of a light bulb if a potential difference of 120 v will produce a current of 0.5 a in the bulb? 0.0042 0.

svlad2 [7]

R = V/I

R = 120 V/ 0.5 A

R = 240 ohms

3 0

3 years ago

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3) A 30 kg child slides freely across a "Slip and Slide" on LEVEL GROUND. While the child slides, the force applied to keep them

zavuch27 [327]

Answer:

a = 0.1962 m/s^2

Explanation:

The magnitude of kinetic friction exerted is given by

$F_k=\mu_kN$

Where, μ_k= coefficient of kinetic friction= 0.02 and N = reaction force = mg

Where m= mass = 30 Kg and, g is acceleration due to gravity =9.81 m/s^2

F_k=0.02×30×9.81 =5.886 N

Now, since, there is no applied force this kinetic friction force will cause acceleration of the child

⇒ ma = F_k

here, a is the acceleration

⇒30a = 5.886

⇒ a = 0.1962 m/s^2

4 0

2 years ago

Brainlist nd 20 point

Alborosie

the same is the answer

6 0

2 years ago

A 14.3-g bullet is fired into a 5.21 kg block of wood. The block is attached to a spring that has a spring force constant of 450

Natasha2012 [34]

Answer:

The initial speed of the bullet is $v_{o} = 889.199\,\frac{m}{s}$ .

Explanation:

The collision between bullet and block is inelastic and let suppose that motion occurs on a horizontal surface, so that changes in gravitational potential energy can be neglected. Initially, the intial speed of the bullet-block system can be determined with the help of the Work-Energy Theorem and the Principle of Energy Conservation:

$K = U_{k} + W_{loss}$

$\frac{1}{2}\cdot (5.224\,kg)\cdot v^{2} = \frac{1}{2}\cdot \left(450\,\frac{N}{m}\right)\cdot (0.22\,m)^{2} + (0.35)\cdot (5.224\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right) \cdot (0.22\,m)$