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3 years ago

What is the materials which disobeys ohm's law?

1 answer:

5 0

Ohm's Law<span> deals with the relationship between voltage and current in an ideal conductor. </span>Ohm's law<span> states that the current through a conductor between two points is directly proportional to the voltage across the two points. </span>

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Bonjour HELPPPP :) LOL

GalinKa [24]

Yeah what the guy above me said

3 0

3 years ago

By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows, with m1 = 12.1 kg and

puteri [66]

Answer:

14.8 kg

Explanation:

We are given that

$m_1=43.7 kg$

$m_2=12.1 kg$

$g=9.8 m/s^2$

$a=\frac{1}{2}(9.8)=4.9 m/s^2$

We have to find the mass of the pulley.

According to question

$T_2-m_2 g=m_2 a$

$T_2=m_2a+m_2g=m_2(a+g)=12.1(9.8+4.9)=177.87 N$

$T_1=m_1(g-a)=43.7(9.8-4.9)=214.13 N$

Moment of inertia of pulley= $I=\frac{1}{2}Mr^2$

$(T_2-T_1)r=I(-\alpha)=\frac{1}{2}Mr^2(\frac{-a}{r})=\frac{1}{2}Mr(-4.9)$

Where $\alpha=\frac{a}{r}$

$(177.87-214.13)=-\frac{1}{2}(4.9)M$

$-36.26=-\frac{1}{2}(4.9)M$

$M=\frac{36.26\times 2}{4.9}=14.8 kg$

Hence, the mass of the pulley=14.8 kg

6 0

3 years ago

When charges build up on a surface they:

NeTakaya

The awnser is A and B

6 0

4 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Svetlanka [38]

The formula written in the 3rd line above the picture is WRONG. Don't use it. Use the formula the way it's printed in the picture.

V = d / t

That means Speed = (distance) / (time)

The question tells us that v = 330 m/s

So you write 330 m/s in the equation in place of 'v', like this:

330 m/s = (distance) / (time)

The question also tells us that the time is 0.4 second

So you write 0.4 sec in place of 'time', like this:

330 m/s = (distance) / (0.4 second)

Finally, you take this, and multiply each side of the equation by (0.4 sec). Then it'll say

distance = (330 m/s) x (0.4 second)

As soon as you do that one single multiplication there with your pencil or your calculator, you'll have the distance.

This is either the 2nd or 3rd time you've posted this same exact question since last weekend. It can be solved THIS time exactly like the answers that were posted those other times.

The DOT in the picture is marked for the wrong choice. Use the formula that's printed in the picture, not copied above it.

6 0

3 years ago

The inner cylinder of a long, cylindrical capacitor has radius r and linear charge density +λ. It is surrounded by a coaxial cyl

Ulleksa [173]

Hi there!

We can begin by using the equation for energy density.

$U = \frac{1}{2}\epsilon_0 E^2$

U = Energy (J)

ε₀ = permittivity of free space

E = electric field (V/m)

First, derive the equation for the electric field using Gauss's Law:
$\Phi _E = \oint E \cdot dA = \frac{Q_{encl}}{\epsilon_0}$

Creating a Gaussian surface being the lateral surface area of a cylinder:
$A = 2\pi rL\\\\E \cdot 2\pi rL = \frac{Q_{encl}}{\epsilon_0}\\\\Q = \lambda L\\\\E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}$

Now, we can calculate the energy density using the equation:
$U = \frac{1}{2} \epsilon_0 E^2$

Plug in the expression for the electric field and solve.

$U = \frac{1}{2}\epsilon_0 (\frac{\lambda}{2\pi r \epsilon_0})^2\\\\U = \frac{\lambda^2}{8\pi^2r^2\epsilon_0}$

Now, we can integrate over the volume with respect to the radius.

Recall:
$V = \pi r^2L \\\\dV = 2\pi rLdr$

Now, we can take the integral of the above expression. Let:
$r_i$ = inner cylinder radius

$r_o$ = outer cylindrical shell inner radius

Total energy-field energy:

$U = \int\limits^{r_o}_{r_i} {U_D} \, dV = \int\limits^{r_o}_{r_i} {2\pi rL *U_D} \, dr$

Plug in the equation for the electric field energy density and solve.

$U = \int\limits^{r_o}_{r_i} {2\pi rL *\frac{\lambda^2}{8\pi^2r^2\epsilon_0}} \, dr\\\\U = \int\limits^{r_o}_{r_i} { L *\frac{\lambda^2}{4\pi r\epsilon_0}} \, dr\\$

Bring constants in front and integrate. Recall the following integration rule:
$\int {\frac{1}{x}} \, dx = ln(x) + C$

Now, we can solve!

$U = \frac{\lambda^2 L}{4\pi \epsilon_0}\int\limits^{r_o}_{r_i} { \frac{1}{r}} \, dr\\\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(r)\left \| {{r_o} \atop {r_i}} \right. \\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} (ln(r_o) - ln(r_i))\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})$

To find the total electric field energy per unit length, we can simply divide by the length, 'L'.

$\frac{U}{L} = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\frac{1}{L} \\\\\frac{U}{L} = \boxed{\frac{\lambda^2 }{4\pi \epsilon_0} ln(\frac{r_o}{r_i})}$

And here's our equation!

3 0

2 years ago