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3 years ago

14

Insulators will:

2 answers:

nalin [4]3 years ago

7 0

Answer:

Inhibit the flow of electrons

Explanation:

An electric current usually consists of electrons moving through a wire.

An insulator prevents the flow of an electric current, so it inhibits the flow of electrons.

Artemon [7]3 years ago

4 0

Answer:

inhibit the flow of electrons

Explanation:

As we know that the electrons are moving from the medium then this type of medium is known as conductors while if the medium will resist the flow of electrons then it is known as insulators

So insulating medium will not allow the electrons to move through the medium

So here all the insulating medium will not allow the flow of electrons while all the conductors will allow the electrons to move through it

All the resistors will allow the flow but also the resist the flow of the charge

so correct answer will be

<u>inhibit the flow of electrons</u>

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In a nuclear fusion reaction, the mass of the products is _____ the mass of the reactants.

JulijaS [17]

Answer:

more than

Explanation:

In a nuclear fusion reaction, the mass of the products is more than the mass of the reactants.

7 0

2 years ago

Can somebody help please !<br><br> a. -8.3 m/s<br> b.-4.2 m/s<br> c.-0.12 m/s<br> d. 0 m/s

Gnom [1K]

The answer is a
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4 0

3 years ago

A block (mass = 61.2 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, wh

kolezko [41]

Answer:

The angular velocity is $w = 53.35 \ rounds /minute$

Explanation:

From the question we are told that

The mass of the block is $m = 61.2kg$

The of the pulley is $M = 14.2 kg$

The radius of the pulley is $R = 1.5m$

The radius of the cord around the pulley is $r = 1.5 m$

The distance of the block to the floor is $d = 8.0 m$

From the question we are told that the moment of inertia of the pulley is

$I = \frac{1}{2} MR^2 kg \cdot m^2$

Substituting value

$I = \frac{1}{2} * 14.2 * (1.5)^2$

$I = 15.975 kg \cdot m^2$

Using the Newtons law we can express the force acting on the vertical axis as

$ma = mg -T$

=> $T = mg -ma$

Now when the pulley is rotated that torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

$\tau = I \alpha$

Here $\alpha$ is the angular acceleration

Here $\tau$ is the torque which can be equivalent to

$\tau = T r$

Substituting this above

$Tr = I \alpha$

Substituting for T

$(mg - ma ) r = I\ r \alpha$

Here $a$ is the linear acceleration which is mathematically represented as

$a = r\alpha$

$(mg - m(r\alpha ) ) r = I\ r \alpha$

$mgr = I\alpha + m(r\alpha ) r$

$mgr = \alpha [ I + mr^2]$

making $\alpha$ the subject

$\alpha = \frac{mgr}{I -mr ^2}$

Substituting values

$\alpha = \frac{61.2 * 1.5 * 9.8}{15.975 + (61.2 ) * (1.5)^2}$

$\alpha =5.854 rad /s^2$

Now substituting into the equation above to obtain the acceleration

$a = 5.854 * 1.5$

$a=8.78 m/s^2$

This acceleration is $a = \frac{v}{t}$

and v is the linear velocity with is mathematically represented as

$v = \frac{d}{t}$

Substituting this into the formula acceleration

$a = \frac{d}{t^2}$

making t the subject

$t = \sqrt{\frac{d}{a} }$

substituting value

$t = \sqrt{\frac{8}{8.78}}$

$t = 0.9545 \ s$

Now the linear velocity is

$v = \frac{8}{0.9545}$

$v = 8.38 m/s$

The angular velocity is

$w = \frac{v}{r}$

So

$w = \frac{8.38}{1.5}$

$w = 5.59 rad/s$

Generally 1 radian is equal to 0.159155 rounds or turns

So 5.59 radian is equal to x

Now x is mathematically obtained as

$x = \frac{5.59 * 0.159155}{1}$

$= 0.8892 \ rounds$

Also

60 second = 1 minute

So 1 second = z

Now z is mathematically obtained as

$z = \frac{ 1}{60}$

z $= 0.01667 \ minute$

Therefore

$w = \frac{0.8892}{0.01667}$

$w = 53.35 \ rounds /minute$

8 0

2 years ago

Can any one give me a amazon code i will repay you. plz i am trying to buy my mom somin off amazon

Alexandra [31]

Answer:i have none but good luck also this is a learning site ;-;

Explanation:

4 0

3 years ago

What is the maximum angular momentum Lmax that an electron with principal quantum number n = 2 can have? Express your answer in

butalik [34]

Answer:

$L_{max} = 1.414$ ℏ

Given:

Principle quantum number, n = 2

Solution:

To calculate the maximum angular momentum, $L_{max}$ , we have:

$L_{max} = \sqrt {l(1 + l)}$ (1)

where,

l = azimuthal quantum number or angular momentum quantum number

Also,

n = 1 + l

2 = 1 + l

l = 1

Now,

Using the value of l = 1 in eqn (1), we get:

$L_{max} = \sqrt {1(1 + 1)} = \sqrt 2$

$L_{max} = 1.414$ ℏ

4 0

3 years ago