Which layer in the image above represents the layer of the Earth composed of liquid metal? *

What are the three reasons cells divide?

hoa [83]

1 growth. Go from one cell/( zygote to a trillion)
2 replace. Repair\ 50 million cells die second.
3 reproduction. ( make cells for reproduction make specialized sex cells)

7 0

4 years ago

Think Critically Suppose your teacher gave you a slide of an unknown cell. How would you tell whether the cell was from an anima

topjm [15]

Answer:

You will be able to tell whether the unknown cell is from an animal or from a plant through the knowledge of difference between plants and animal cells.

Explanation:

A cell can be defined as the simplest bit of living matter which exhibits a variety of shapes and sizes and that can exist independently.

When a slide of an unknown cell is studied under a microscope, different cell structures would be observed which would be used to conclude if the cell on the slide is a plant or animal cell.

The following features are observed in a plant cell:

--> cellulose cell wall is present

--> Chloroplast is present

--> A large central vacuole is present

--> Centriole is absent

The following features are observed in animal cell:

--> Cellulose cell wall is absent

--> Chloroplast is absent

--> Small vacuoles is present

--> Centriole is present

The difference between a plant and an animal cell can be seen from the features above and a clear knowledge of this will help the student tell whether the unknown cell is from an animal or from a plant.

4 0

3 years ago

A cube made Of an unknown material has a height of 9 symeters the mass of this cube is 3645 g. calculate the density of this cub

kherson [118]

Answer: $5 \frac{g}{cm^{3}}$

Explanation:

The density $\rho$ of a material is given by:

$\rho=\frac{m}{V}$ (1)

Where:

$m=3645 g$ is the mass of the cube

$V$ is the volume of the cube

Now, the volume of a cube is equal to the length $L$ of its edge to the power of 3:

$V=L^{3}$ (2)

If we know $L=9 cm$ , the volume of this cube is:

$V=(9 cm)^{3}=729 cm^{3}$ (3)

Substituting (3) in (1):

$\rho=\frac{3645 g}{729 cm^{3}}$ (4)

$\rho=5 \frac{g}{cm^{3}}$ This is the density of the cube

8 0

4 years ago

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

raketka [301]

Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

Projectile is thrown with a velocity = v
Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

$\qquad$ ______________________________

Then –

$\qquad$ $\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\qquad$ $\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be –

$\qquad$ $\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }$

$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

6 0

3 years ago

Read 2 more answers

If you ride quickly down a hill on a bicycle your eardrums are pushed in before they pop back. Why is this?

Mice21 [21]

Answer:

<em>The difference in pressure between the external air pressure, and the internal air pressure of the middle ear.</em>

Explanation:

First of all, we should note that pressure decreases with height and increases with depth. The air within the middle ear (between the ear drum and the Eustachian tube) adjusts itself to respond to the atmospheric pressure, or when we yawn. At a high altitude like on the hill, the air pressure in the middle ear, is fairly low (this is to balance the low air pressure at this height). While riding down the hill quickly, there is little time for the air pressure in the ear to readjust itself to the increasing external air pressure, causing the external air to push into the ear drum. Along the way, the air within the middle ear is adjusted by the opening of the Eustachian tube, allowing more air into the space in the middle ear to balance the external air pressure. This readjustment causes the ear to pop.

7 0

3 years ago