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zhenek [66]
3 years ago
8

Why is it necessary for cells to be so small​

Physics
1 answer:
madreJ [45]3 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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A frog leaping upward off his lily pad instead of continuing on in a straight line . Which law is it ?
Romashka [77]

Answer:

I think it's Law of Motion.

Explanation:

Hope my answer has helped you!

7 0
3 years ago
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Which of these is a process during which the hydrosphere interacts with the geosphere?
lana66690 [7]
<span>Answer: b. absorption of rainwater by soil

The word hydro in hydrosphere means water and the word geo in geosphere mean the earth soil. The system of hydrosphere will interact when the soil adsorption the rainwater. The other options are just interaction of hydrosphere without geosphere.</span>
7 0
4 years ago
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What type of energy is thermal energy? A. Sound energy B. Chemical energy C. Kinetic energy D. Nuclear energy
Nadya [2.5K]

Answer:

kinetic energy

Explanation:

7 0
3 years ago
a force 2.4E2 N exists between a positive charge of 8E-5 C and a positive charge of 3E-5 C. What distance separates the charges?
Verdich [7]

The distance between the two charges is 0.3 m

Explanation:

The electrostatic force between two charged objects is given by Coulomb's law:

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

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the charges of the two objects

r is the separation between the two charges

In this problem, we are given the following:

q_1 = 8\cdot 10^{-5} C

q_2 = 3\cdot 10^{-5} C

F=2.4\cdot 10^2 N

Therefore, we can rearrange the equation to solve for r, the distance between the two charges:

r=\sqrt{\frac{kq_1 q_2}{F}}=\sqrt{\frac{(8.99\cdot 10^9)(8\cdot 10^{-5})(3\cdot 10^{-5})}{2.4\cdot 10^2}}=0.3 m

Learn more about electrostatic force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0
3 years ago
The speed of an arrow fired from a compound
san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

X=v_{ox}*t

v_{0x} =v_0cos(\alpha)

Y axis:

Y= Y_0 +v_{y0} t - \frac{g}{2} t^2

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, Y=0, then

0= Y_0 +v_{y0} t - \frac{g}{2} t^2

0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2

Caculate the segcond degree equation to obtain the two possible values for t:

t_1= 10.18 \\t_2= -0.04046

But, in physics, time it could not be negative, so we take t_1= 10.18

Caculate now:

X=79m/s*cos(\39)*10.18s= 624.996 m

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

v_0= 79m/s+13m/s= 92m/s

Using the same procedure that item A, caculate X

First, we need to know the new time

0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2

And we obtain:

t_1=11.845s\\t_2=-0.041s

One more time, we take the positive time: t_1=11.845s

Finally:

X=92m/s *cos(39)*11.845s=846.887 m

6 0
3 years ago
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