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3 years ago

14

An electrician wraps rubber electrical tape around a copper wire. Since rubber is an insulator, which statement best explains th

e point of wrapping rubber around a copper wire?

1 answer:

yuradex [85]3 years ago

4 0

A) Rubber stops charges from flowing. This protects people by stopping electricity from flowing.

Explanation:

The statement that best describes the point of wrapping rubber around the copper wire is that the rubber stops charges from flowing. This prevents people from getting electrical shocks by stopping the flow of electricity.

A rubber is an insulator.
Insulators are substances that prevents the flow of electricity.
The lack free mobile electrons or ions that makes them conductors.
When they are wrapped round a conductor such as copper wire, they will halt the flow of charges.
Copper is a conductor of both heat and electricity. It has free mobile electrons.

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You might be interested in

efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.

PolarNik [594]

Answer:

Inlet : $v_i=0.0646\frac{m}{s}$

Outlet: $v_o=0.171\frac{m}{s}$

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

$\dot{m}=\frac{\dot{V}}{\upsilon}$ , where $\dot{V}$ represent the flow rate and $\upsilon$ the specific volume at the pressure and temperature given.

$A_i=0.5m^2$ is the inlet area

$P_i=600Kpa$ pressure at the inlet area

$T_i=70C$ temperature at the inlet area

$A_o=1m^2$ is the outlet area

$P_o=100Kpa$ pressure at the outlet area

$T_o=C$ temperature at the outlet area

$\dot{m}=0.75\frac{kg}{s}$ represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

$\upsilon_i =0.04304\frac{kg}{m^3}$ and the entropy is $h_i=1.0645\frac{KJ}{KgK}=h_o$

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy $h_o=1.0645\frac{KJ}{KgK}$

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

$h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}$

$h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}$

Our interest value would be given using interpolation like this:

$\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}$

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that $\dot{m}=\frac{\dot{V}}{\upsilon}$ , but since $\dot{V}=Av$ we have this:

$\dot{m}=\frac{Av}{\upsilon}$

If we solve from the velocity v we have this:

$v=\frac{\upsilon \dot{m}}{A}$ (*)

And now we just need to replace the values into equation (*)

For the inlet case:

$v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}$

For the oulet case:

$v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}$

7 0

3 years ago

The near point of an eye is 56.0 cm. A corrective lens is to be used to allow this eye to focus clearly on objects at the distan

irakobra [83]

Answer:

Explanation:

Near point = 56 cm .

near point of healthy person = 25 cm

person suffers from long sightedness

convex lens will be required .

object distance u = 25 cm

image distance v = 56 cm

both will be negative as both are in front of the lens.

lens formula

I/v - 1 / u = 1/f

- 1/56 +1/25 = 1/f

- .01785 + .04 = 1/f

1/f = .02215

f = 45.15 cm .

4 0

3 years ago

Two lightbulbs both operate on 120V . One has a power of 25W and the other 100W. (ii) Which lightbulb carries more current? Choo

Vikki [24]

The lightbulb that carries more current will be the 25W bulb.

<h3>How to explain the information?</h3>

It should be noted that an electric current simply means the stream of charged particles that move through an electrical conductor or space.

In this case, it should be noted that the power is the same for both bulbs. Therefore, the 25W bulb will have the higher resistance so that it will have lower power.

Therefore, the lightbulb that carries more current will be the 25W bulb.

Learn more about current on:

brainly.com/question/1100341

#SPJ4

7 0

1 year ago

The temperature of a 700.96 gram piece of metal falls 120⁰C and in the process releases 2001 Joules of energy. What is the speci

olga nikolaevna [1]

Answer:

The specific heat for the metal is 0.466 J/g°C.

Explanation:

Given,

Q = 1120 Joules

mass = 12 grams

T₁ = 100°C

T₂ = 300°C

The specific heat for the metal can be calculated by using the formula

Q = (mass) (ΔT) (Cp)

ΔT = T₂ - T₁ = 300°C - 100°C = 200°C

Substituting values,

1120 = (12)(200)(Cp)

Cp = 0.466 J/g°C.

Therefore, specific heat of the metal is 0.466 J/g°C.

8 0

3 years ago

What magnet maintains its magnetic proaperties even in the absence of an external magnetic field? a) Ferromagnet b Paramagnet c)

insens350 [35]

Answer:

(a) Ferromagnet

Explanation:

Ferromagnetism is defined as the property by which certain magnets form the permanent magnets.

It is tone of the strong magnetism and it is common phenomenon of magnet in the everyday life of magnetism.

Permanent magnets are made up of ferromagnetic material, in this if the magnetic field is applied then this material is magnetized but do not losses its magnetic property after removal of external magnetic field.

5 0

3 years ago