All categories

3 years ago

Like magnetism, static electricity can attract and repel. True False

Physics

2 answers:

earnstyle [38]3 years ago

5 0

The answer to the question is True

sweet-ann [11.9K]3 years ago

4 0

True because my hair demonstrates it xD

You might be interested in

The solid right-circular cylinder of mass 500 kg is set into torque-free motion with its symmetry axis initially aligned with th

aivan3 [116]

Answer:

30.95°

Explanation:

We need to define the moment of inertia of cylinder but in terms of mass, that equation say,

$A=\frac{1}{12}m(3r^2+l^2)$

Replacing the values we have,

$A=\frac{1}{12}(500)(3(0.5)^2+(2)^2)$ $A=197.9kg.m^2$

At the same time we can calculate the mass moment of intertia of cylinder but in an axial way, that is,

$c=\frac{1}{2}mr^2$

$c=\frac{1}{2}(500)(0.5)^2$

$c=62.5kg.m^2$

Finally we need to find the required angle between the fixed line a-a (I attached an image )

$\Phi = 2tan^{-1}\sqrt{\frac{(\frac{A}{cos\gamma})^2-A^2}{c^2}}$

Replacing the values that we have,

$\Phi = 2tan^{-1}\sqrt{\frac{(\frac{197.9}{cos5\°})^2-197.9^2}{62.5^2}}$

$\Phi = 2tan^{-1}(\sqrt{0.076634})$

$\Phi = 2tan^{-1}(0.2768)$

$\Phi = 2(15.47)$

$\Phi = 30.95\°$

4 0

4 years ago

Please help me! It’s due soon! Please help!

Murrr4er [49]

Answer:

do it yourself lazy h.e

Explanation:

kysc lazy btc

4 0

3 years ago

It usually takes more force to start an object sliding than it does to keep an object sliding because static friction is usually

labwork [276]

The answer is already in the blank for, its was greater

5 0

4 years ago

Suppose that a spiral galaxy is located at the center of a spherically symmetric dark matter halo

Novosadov [1.4K]

To solve the problem it is necessary to use the concepts of Orbital Speed considering its density, and its angular displacement.

In general terms the Orbital speed is described as,

$V_{orbit} = \sqrt{\frac{G\rho 4\pi r^3}{3}}$

PART A) If the orbital speed of a star in this galaxy is constant at any radius, then,

$\frac{4\pi G\rho r}{3} = \frac{v^2}{r}$

$\frac{4\pi G\rho r}{1} = \frac{3v^2}{r}$

$\frac{\rho}{1} = \frac{3v^2}{r^2 4\pi G}$

$\rho = \frac{1}{r^2}$

PART B) This time we have $v=\omega t$ , where $\omega$ is the angular velocity (constant at this case)

$\frac{4\pi G\rho r}{3} = \frac{v^2}{r}$

$\frac{4\pi G\rho r}{3} = \frac{(\omega r)^2}{r}$

$\rho = \frac{3\omega r}{4\pi Gr}$

$\rho = \frac{3\omega^2}{4\pi G} \propto constant$

PART C) If the total mass interior to any radius r is a constant,

$\frac{4\pi G\rho r}{3} = \frac{v^2}{r}$

$\frac{GM}{r^2}=\frac{v^2}{r}$

$v = \sqrt{\frac{GM}{r}}$

$v= \sqrt{\frac{1}{r}}$

3 0

4 years ago

A coil of wire containing N turns is in an external magnetic field that is perpendicular to the plane of the coil and it steadil

krok68 [10]

Answer:

The Resultant Induced Emf in coil is 4∈.

Explanation:

Given that,

A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.

To find :-

find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).

So,

Emf induced in the coil represented by formula

∈ = $-N\frac{d\phi}{dt}$ ...................(1)

Where:

. $\phi = BAcos\theta$ { B is magnetic field }

{A is cross-sectional area}

. $N =$ No. of turns in coil.

. $\frac{d\phi}{dt} =$ Rate change of induced Emf.

Here,

Considering the case :-

$N1 = 2N$ & $\frac{d\phi1}{dt} = 2\frac{d\phi}{dt}$

Putting these value in the equation (1) and finding the new emf induced (∈1)

∈1 = $-N1\times\frac{d\phi1}{dt}$

∈1 = $-2N\times2\frac{d\phi}{dt}$

∈1 = $4 [-N\times\frac{d\phi}{dt}]$

∈1 = 4∈ ...............{from Equation (1)}

Hence,

The Resultant Induced Emf in coil is 4∈.

8 0

4 years ago