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3 years ago

What kind of energy does each phase of the pendulum represent

Physics

2 answers:

adelina 88 [10]3 years ago

3 0

Answer:

Kinetic

Explanation:

Potienal is building up power where kinetic means moving

ElenaW [278]3 years ago

3 0

Answer:

Position A and Position C = Potential energy

Position B = Kinetic energy

Explanation:

When pendulum moves to and fro about its mean position then the velocity of pendulum is maximum at its mean position or at its lowest position of path.

As it move away from its lowest position the velocity of pendulum keep on decreasing and at the end it comes to rest.

So at the extreme ends of the path the pendulum will come to rest so its kinetic energy is zero and all its energy will become potential energy.

So here we know that in the given figure position A and Position C is extreme positions and the energy of that position is potential energy.

While at the lowest position all its energy becomes kinetic energy

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Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0

3 years ago

The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A

kotykmax [81]

Answer:

$2.1\times 10^{-12} c$

Explanation:

We are given that

Surface area of membrane= $5.3\times 10^{-9} m^2$

Thickness of membrane= $1.1\times 10^{-8} m$

Assume that membrane behave like a parallel plate capacitor.

Dielectric constant=5.9

Potential difference between surfaces=85.9 mV

We have to find the charge resides on the outer surface of membrane.

Capacitance between parallel plate capacitor is given by

$C=\frac{k\epsilon_0 A}{d}$

Substitute the values then we get

Capacitance between parallel plate capacitor= $\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}$

$C=0.25\times 10^{-12}F$

V= $85.9 mV=85.9\times 10^{-3}$

$Q=CV$

$Q=0.25\times 10^{-12}\times 85.9\times 10^{3}=2.1\times 10^{-12} c$

Hence, the charge resides on the outer surface= $2.1\times 10^{-12} c$

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3 years ago

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If I were a geologist working in South America along one of those plate boundaries, what might I see if I were looking at sedime

faust18 [17]

Answer:

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Rina8888 [55]

Answer:

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A Porsche will always win no matter what

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