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3 years ago

Paragraph about cell wall

2 answers:

7 0

Animal cells do not have cell walls, nor do protozoa. Cell walls protect the cells from damage. It is also there to make the cell strong, to keep its shape, and to control the growing of the cell and plant. The cell wall is the tough, usually flexible but sometimes fairly rigid layer that surrounds some types of cells

ELEN [110]3 years ago

7 0

All you need to know about the cell wall is that it protects the cells inside of it.

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From January 26, 1977, to September 18, 1983, George Meegan of Great Britain walked from Ushuaia, at the southern tip of South A

Nikolay [14]

Answer:

0.146 m/s

Explanation:

We can see it in the pic.

3 0

3 years ago

1. If a car travels 400m in 20 seconds how fast is it going?

Marrrta [24]

Answer:

1) 20 m/s

2) 5 m/s

3) 2 m/s

4) 395,000m/9000s

5) 16 km/0.25h

Explanation:

i dunno

5 0

3 years ago

Which statements correctly describe the properties of air? Select three options. Air has mass but no weight. Air pressure is cre

jolli1 [7]

Answer:

2,4,5

Air pressure is created by the weight of the atmosphere pushing on Earth’s surface.

Denser air is heavier than less dense air.

Air is less dense at higher altitudes.

4 0

3 years ago

Read 2 more answers

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

4 0

3 years ago

To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeof

il63 [147K]

<h2> The potential and kinetic energy of airplane are affected by these factors </h2>

Explanation:

When airplane rises up , it requires potential energy . This potential energy can be taken from the kinetic energy of airplane .

Thus if the speed of wind is larger , it can either oppose the motion of velocity or can favour the velocity of airplane . By which its kinetic energy is effected .

If the weight of airplane is changed , it will effect the potential energy required . Thus heavier plane requires higher potential energy for attaining the same height .

Thus these two factor has important role in the flight of airplane .

6 0

4 years ago