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zhenek [66]
2 years ago
8

Why is it necessary for cells to be so small​

Physics
1 answer:
madreJ [45]2 years ago
8 0

Answer:

Cells are small because they need to keep a surface area to volume ratio that allows for adequate intake of nutrients while being able to excrete the cells waste.

Explanation:

That is why the cell needs to be small

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Describe what a planetesimal is. Is it related to a protoplanet?
zheka24 [161]

Answer:

A planetesimal is an object formed from dust, rock, and other materials. The word has its roots in the concept infinitesimal, which indicates an object too small to see or measure. Planetesimals can be anywhere in size from several meters to hundreds of kilometers.

Protoplanets are thought to form out of kilometer-sized planetesimals that gravitationally perturb each other's orbits and collide, gradually coalescing into the dominant planets.

4 0
3 years ago
What is the final concentration of DD at equilibrium if the initial concentrations are [A][A]A_i = 1.00 MM and [B][B]B_i = 2.00
pentagon [3]

Answer:

A) Concentration of A left at equilibrium of we started the reaction with [A] = 2.00 M and [B] = 2.00 M is 0.55 M.

B) Final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M is 0.90 M.

[D] = 0.90 M

Explanation:

With the first assumption that the volume of reacting mixture doesn't change throughout the reaction.

This allows us to use concentration in mol/L interchangeably with number of moles in stoichiometric calculations.

- The first attached image contains the correct question.

- The solution to part A is presented in the second attached image.

- The solution to part B is presented in the third attached image.

8 0
3 years ago
Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu
lora16 [44]

The initial force between the two charges is given by:

F=k \frac{q_1 q_2}{d^2}

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

q_1' = q_1\\q_2' = q_2\\d' = 2d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F

So the force has increased by a factor 6.

8 0
2 years ago
Read 2 more answers
Honeybees can see light in the ________ range of the electromagnetic spectrum. question 8 options:
ELEN [110]
<span>ultraviolet

Have a great day!</span>
8 0
3 years ago
Read 2 more answers
Superman does an exhibition run at a track meet. When he runs the 200 m
Ber [7]

Answer:

6.32s

Explanation:

Given parameters:

Length of track and distance covered  = 200m

Acceleration  = 10m/s²

Unknown:

Time taken to cover the track  = ?

Solution:

To solve this problem, we apply one of the motion equations as shown below:

       S  = ut + \frac{1}{2} at²  

S is the distance covered

t is the time taken

a the acceleration

u is the initial velocity

The initial velocity of Superman is 0;

 So;

     S  =  \frac{1}{2} at²  

        200  =  \frac{1}{2} x 10 x t²  

          200  = 5t²  

            t²  = 40

            t  = 6.32s

5 0
3 years ago
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