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just olya [345]
3 years ago
6

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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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:

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3 years ago
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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.

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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

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Here,

C = Capacitance

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Q = Charge between the capacitor plates

At the same time the energy stored in the capacitor can be defined as,

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We will start by finding the value of the capacitance, so we will have to,

C = \frac{2\mu C}{12.0V}

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Finally using the expression for the energy we have that,

U = \frac{1}{2} CV^2

U = \frac{1}{2} (0.166\muF)(12.0V)^2

U = 11.0*10^{-6} J

Therefore the minimum amount of work that must be done in charging this capacitor is 11.0*10^{-6} J

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