Describe how energy is transformed in the motor

A capacitor has plates of area 8.25 * 10 ^ - 5 m ^ 2 . To create a capacitance of 3.35*10^ -10 F , how far apart should the plat

sveticcg [70]

Answer:

A capacitor has plates of area 8.25 * 10 ^ - 5 m ^ 2 . To create a capacitance of 3.35*10^ -10 F , how far apart should the plates be ?

Explanation:

6 0

2 years ago

Chloroplasts convert energy from the sun into ________________.Immersive Reader

Aleksandr [31]

Answer:

1) glucose

2) ATP

Explanation:

moo

6 0

3 years ago

As electric current moves through a wire, heat generated by resistance is conducted through a layer of insulation and then conve

mestny [16]

Answer:

Hello your question is incomplete attached below is the complete question

answer : ri ( thickness of wire ) = 14.167 m

Explanation:

attached below is a detailed solution

using the given data to determine the thickness of wire

5 0

2 years ago

A machine part has the shape of a solid uniform sphere of mass 220 g and diameter 4.50 cm . It is spinning about a frictionless

miss Akunina [59]

Answer:

The angular acceleration is 10.10 rad/s².

Explanation:

Given that,

Mass of sphere =220 g

Diameter = 4.50 cm

Friction force = 0.0200 N

Suppose we need to find its angular acceleration.

We need to calculate the angular acceleration

Using formula of torque

$\tau=f\times r$

$I\times\alpha=f\times r$

Here, I = moment of inertia of sphere

$\dfrac{2}{5}mr^2\times\alpha=f\times r$

$\alpha=\dfrac{5\times f}{2mr}$

Put the value into the formula

$\alpha=\dfrac{5\times0.0200}{2\times220\times10^{-3}\times2.25\times10^{-2}}$

$\alpha=10.10\ rad/s^2$

Hence, The angular acceleration is 10.10 rad/s².

7 0

3 years ago

A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words: