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3 years ago

The steps of mieosis

Physics

2 answers:

Grace [21]3 years ago

8 0

Answer:

here are six stages within each of the divisions, namely prophase, prometaphase, metaphase, anaphase, telophase and cytokinesis

Explanation:

lbvjy [14]3 years ago

4 0

Answer:

Since cell division occurs twice during meiosis, one starting cell can produce four gametes (eggs or sperm). In each round of division, cells go through four stages: prophase, metaphase, anaphase, and telophase.

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What would be the projectile's initial speed?

zalisa [80]

Math is the process of using the given information, along with
all the general stuff that you know, to find the missing information.
With no given information, we have no way to even guess at
an answer.

8 0

4 years ago

Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor

fenix001 [56]

Answer:

a) $F_{r}= -583.72MN i + 183.47MN j + 6.05GN k$

b) $E=3.04 \frac{GN}{C}$

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

$r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}$

which yields:

$r_{AP}^{2}= 43 m^{2}$

(I will assume the positions are in meters)

Next, we can make use of the force formula:

$F=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

so we substitute the values:

$F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}$

which yields:

$F_{AP}=418.14 MN$

Now we can find its components:

$F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i$

$F_{APx}=191.30 MNi$

$F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j$

$F_{APy}=191.30MN j$

$F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k$

$F_{APz}=318.83 MN k$

And we can now write them together for the first force, so we get:

$F_{AP}=(191.30i+191.30j+318.83k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}$

which yields:

$r_{BP}^{2}= 33 m^{2}$

Next, we can make use of the force formula:

$F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}$

which yields:

$F_{BP}=2.18 GN$

Now we can find its components:

$F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i$

$F_{BPx}=758.98 MNi$

$F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j$

$F_{BPy}=758.98MN j$

$F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k$

$F_{BPz}=1.897 GN k$

And we can now write them together for the second, so we get:

$F_{BP}=(758.98i + 758.98j + 1897k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}$

which yields:

$r_{CP}^{2}= 30 m^{2}$

Next, we can make use of the force formula:

$F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}$

which yields:

$F_{CP}=4.20 GN$

Now we can find its components:

$F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i$

$F_{CPx}=-1.534 GNi$

$F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j$

$F_{CPy}=-766.81 MN j$

$F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k$

$F_{CPz}=3.83 GN k$

And we can now write them together for the third force, so we get:

$F_{CP}=(-1.534i - 0.76681j +3.83k)GN$

So in order to find the resultant force, we need to add the forces together:

$F_{r}=F_{AP}+F_{BP}+F_{CP}$

so we get:

$F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN$

So when adding the problem together we get that:

$F_{r}=(-0.583.72i + 0.18347j +6.05k)GN$

which is the answer to part a), now let's take a look at part b).

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

$F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}$

which yields:

$F_{r}=6.08 GN$

and now we take the formula for the electric field which is:

$E=\frac{F_{r}}{q}$

so we go ahead and substitute:

$E=\frac{6.08GN}{2C}$

$E=3.04\frac{GN}{C}$

7 0

4 years ago

If bert the bat traveles eastward at 40 mph with a tail wind of 6 mph,what is his actual speed?

Liula [17]

Bert would be going 34mph due to the taill winds being 6mph

5 0

3 years ago

Study the velocity vs. time graph shown.

laiz [17]

For velocity vs Time graphs, the displacement of the object from 2 seconds to 6 seconds is 30 m.

<h3>What is displacement?</h3>

The displacement is the shortest distance travelled by the particle. It is the vector quantity which re[presents both the magnitude and direction.

In velocity time graphs, the displacement is the area under the curve of the graph on the x axis.

A line starts at (0, 2) and ends at (6, 8) in v-t graph

Displacement is equal to the area of a triangle and a rectangle formed under the line.

Area = 1/2 base x height + length x breadth

Area = 1/2 x 6x 6 + 6x2

Area = 18 +12

Area = 30 m

Thus, the displacement is 30 m.

Learn more about displacement.

brainly.com/question/11934397

#SPJ1

6 0

2 years ago

a parallel plate capacitor has square plates that have edge length equal to 100cm and are separated by 1 mm. It is connected to

Flura [38]

Answer:

the energy is stored in the capacitor is 0.32 μJ

Explanation:

Given;

distance of separation, d = 1 mm = 0.001 m

edge length of the square = 100 cm

potential difference across the plates, V = 12 v

let the side of the square = L

This edge length is also the diagonal of the square which makes a right angle with the side of the square.

Applying Pythagoras theorem;

L² + L² = 100²

2L² = 100²

L² = 100²/2

Note area of a square is L²

A = L² = 100²/2 = 5000 cm²

A (m²) = 5000 cm² x 1m²/(100 cm)²

A = 5000 cm² x 1m²/10000 cm²

A = 0.5 m²

Energy stored in a parallel plate capacitor, E= ¹/₂CV²

C = ε₀A/d

where;

ε₀ is permittivity of free space = 8.85 x 10⁻¹² F/m

d is the distance of separation = 0.001 m

A is the area of the plate

C = ε₀A/d = (8.85 x 10⁻¹²)x0.5 / 0.001

C = 4425 x 10⁻¹² F

E = ¹/₂CV² = ¹/₂ x 4425 x 10⁻¹² x ( 12)²

E = 318600 x 10⁻¹² = 0.32 μJ

Therefore, the energy is stored in the capacitor is 0.32 μJ

8 0

3 years ago

Read 2 more answers