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3 years ago

When an object gets larger, why does the volume of an object increase faster than the surface area?

Physics

1 answer:

alina1380 [7]3 years ago

6 0

Let us consider the object to be a cell.

Explanation:

As a cell grows bigger, its internal volume enlarges and the cell membrane expands. This results in the increase in volume more rapidly than does the surface area, and so the relative amount of surface area available to pass materials to a unit volume of the cell steadily decreases. This means the surface area to the volume ratio gets smaller as the cell gets larger.

Example of a cube:

Cube size Surface area Volume

2cm 2 × 2 × 6 = 24 cm² 2 × 2 × 2 = 8 cm³

4 cm 4 × 4 × 6 = 96 cm² 4 × 4 × 4 = 64 cm³

6 cm 6 × 6 × 6 = 216 cm² 6 × 6 × 6 = 216 cm³

8 cm 8 × 8 × 6 = 384 cm² 8 × 8 × 8 = 512 cm³

This shows as the object gets larger, the volume of an object increases faster than the surface area.

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How does Newton’s first second and third laws apply to eating your breakfast

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Answer:

Newton's third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A. Notice that the forces are exerted on different objects. However much you push your fork in the food, that much a dent it will cause.

Explanation:

6 0

3 years ago

Read 2 more answers

A 1.10 μF capacitor is connected in series with a 1.92 μF capacitor. The 1.10 μF capacitor carries a charge of +10.1 μC on one p

miss Akunina [59]

Answer:

(a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.

Explanation:

Given that,

Charge = 10.1 μC

Capacitor C₁ = 1.10 μF

Capacitor C₂ = 1.92 μF

Capacitor C₃ = 1.10 μF

Potential V₁ = 51.5 V

Let V₁ and V₂ be the potentials on the two plates of the capacitor.

(a). We need to calculate the potential on the negative plate of the 1.10 μF capacitor

Using formula of potential difference

$V_{1}=\dfrac{Q}{C_{1}}$

Put the value into the formula

$V_{1}=\dfrac{10.1 \times10^{-6}}{1.10\times10^{-6}}$

$V_{1}=9.18\ V$

The potential on the second plate

$V_{2}=V-V_{1}$

$V_{2}=51.5 -9.18$

$V_{2}=42.32\ v$

(b). We need to calculate the equivalent capacitance of the two capacitors

Using formula of equivalent capacitance

$C=\dfrac{C_{1}\timesC_{2}}{C_{1}+C_{2}}$

Put the value into the formula

$C=\dfrac{1.10\times10^{-6}\times1.92\times10^{-6}}{(1.10+1.92)\times10^{-6}}$

$C=6.99\times10^{-7}\ F$

$C=0.69\ \mu F$

Hence, (a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.

7 0

3 years ago

According to the reading, which of the following is NOT an obstacle to using big data for analysis and decision making? O Poor r

RoseWind [281]

Answer:

B .Limited types of data Insufficient data

Explanation:

To analyse and making decision from big data we have following thing

- Sufficient data

- data analyst

- Accurate data

- data Privacy

- data storage

So, limited data or data that cannot be copied is not and obstacle in data handling. hence option B Limited types of data Insufficient data is correct.

3 0

3 years ago

Is a dog barking an example of kinetic energy to potential energy

Pepsi [2]

It is not an example of kinetic to potential because the dog is already in motion rather then having the potential to do something. Hope this helps!

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2 years ago

A charge of 2.00 μC flows onto the plates of a capacitor when it is connected to a 12.0-V potential source. What is the minimum

uysha [10]

To develop this problem we will apply the concept of energy conservation. For which the work carried out must be equivalent to the potential energy stored on the capacitor. We will start by finding the capacitance to later be able to calculate the energy and therefore the work in the capacitor

$C = \frac{Q}{V}$