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3 years ago

6

What is the process of cell division that results in the creation of four sex cells? A. budding B. fertilization C. meiosis D. m

itosish

2 answers:

babymother [125]3 years ago

7 0

Answer:

meiosis

Explanation:

meiosis produces four sex cells

pentagon [3]3 years ago

4 0

Meiosis is a type of cell division that reduces the number of chromosomes in the parent cell by half and produces four gamete cells. This process is required to produce egg and sperm cells for sexual reproduction. ... Meiosis begins with a parent cell that is diploid, meaning it has two copies of each chromosome.

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Schach [20]

Answer:

how close a true value a measurement is

6 0

3 years ago

Read 2 more answers

A satellite in the shape of a solid sphere of mass 1,900 kg and radius 4.6 m is spinning about an axis through its center of mas

algol [13]

Answer:

6.3 rev/s

Explanation:

The new rotation rate of the satellite can be found by conservation of the angular momentum (L):

$L_{i} = L_{f}$

$I_{i}*\omega_{i} = I_{f}*\omega_{f}$

The initial moment of inertia of the satellite (a solid sphere) is given by:

$I_{i} = \frac{2}{5}m_{s}r^{2}$

Where $m_{s}$ : is the satellite mass and r: is the satellite's radium

$I_{i} = \frac{2}{5}m_{s}r^{2} = \frac{2}{5}1900 kg*(4.6 m)^{2} = 1.61 \cdot 10^{4} kg*m^{2}$

Now, the final moment of inertia is given by the satellite and the antennas (rod):

$I_{f} = I_{i} + 2*I_{a} = 1.61 \cdot 10^{4} kg*m^{2} + 2*\frac{1}{3}m_{a}l^{2}$

Where $m_{a}$ : is the antenna's mass and l: is the lenght of the antenna

$I_{f} = 1.61 \cdot 10^{4} kg*m^{2} + 2*\frac{1}{3}150.0 kg*(6.6 m)^{2} = 2.05 \cdot 10^{4} kg*m^{2}$

So, the new rotation rate of the satellite is:

$I_{i}*\omega_{i} = I_{f}*\omega_{f}$

$\omega_{f} = \frac{I_{i}*\omega_{i}}{I_{f}} = \frac{1.61 \cdot 10^{4} kg*m^{2}*8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kg*m^{2}} = 6.3 rev/s$

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

I hope it helps you!

5 0

4 years ago

A small aircraft accelerated down a runway at 4.0 m/s²

Radda [10]

Given data in the problem :-

Acceleration (a) = 4.0 m/s^2
Initial velocity (u) = 0 m/s
Final velocity (v) = 34 m/s
Distance travelled by aircraft (S) = ?

From Newton's Laws of Motion we know that ,

v = u + at [t = Time taken by aircraft to cover the distance]

⇒ 34 = 0 + 4t

⇒ t = 34/4 s

∴ t = 8.5 s

From Newton's Laws of Motion we also know that ,

S = u.t + 1/2a.t^2

⇒ S = 0×8.5 + 1/2 × 4 × (8.5)^2 m

∴ S = 144.50 m

Thus the distance travelled by the aircraft while accelerating is 144.50 meter .

4 0

2 years ago

A ball is thrown vertically upwards with a velocity v and an initial kinetic energy Ek. When half way to the top of its flight,

Mama L [17]

Answer:

Option A

$\frac{V}{\sqrt{2} }$ , $\frac{E}{2}$

Explanation:

When the ball its thrown up, at half way of its flight it means half of its vertical height which is $\frac{h}{2}$ .

potential energy = mgh

since it moved half way of height

P.E = $\frac{mgh}{2}$

This means for the body to have gained half of its P.E, it will loose half of its kinetic energy.

Final kinetic energy( $E_{1}$ ) = E/2

kinetic energy = $\frac{1}{2}mv^{2}$

let the final velocity at halfway flight be v1

$E_{1}$ = E/2

$\frac{1}{2}mv_{1}^{2}$ = $\frac{\frac{1}{2}mv^{2}}{2}$

cross multiply we have

$mv_{1}^{2}$ = $\frac{1}{2}mv^{2}$

cancel m from both sides

$v_{1}^{2}$ = $\frac{1}{2}v^{2}$

take the square root of both sides,

$v_{1} =\sqrt{\frac{v^{2} }{2} }$

$v_{1} =\frac{v}{\sqrt{2} }$

Thus our final velocities will be E/2 and $\frac{v}{\sqrt{2} }$

5 0

3 years ago

How do mass and distance affect the gravitational force between two objects?

Dafna1 [17]

Answer:

The force of gravity depends directly upon the masses of the two objects and inversely on the square of the distance between them.This means that the force of gravity increases with mass but decreases with increasing distance between objects

8 0

3 years ago