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3 years ago

What is required for an electric charge to flow through a wire?

2 answers:

max2010maxim [7]3 years ago

5 0

In order to persuade the electrons in the wire to flow, you need
a potential difference between the ends of the wire. Then the
electrons will want to get away from the more-negative end and
go to the more-positive end. If both ends of the wire are at the
same potential, then the electrons have no reason to go anywhere,
and they just stay where they are.

Choice-d says this.

Fudgin [204]3 years ago

5 0

The correct answer is:

D. A difference in electric potential.

A difference in electric potential is required for an electric charge to flow through a wire.

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What are the main characteristics of S.H.M

agasfer [191]

Answer:

•→ The motion of a particle or body in S.H.M acts towards a fixed point.

•→ Acceleration of the body under S.H.M is proportional to its displacement.

•→ This motion is periodic.

•→ Mechanical energy is conserved in S.H.M

Explanation:

S.H.M is Simple Harmonic Motion

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2 years ago

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State advantages of ultrasonic sound in determining the depth of the ocean

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Explanation:

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1 year ago

Describes the relationship between the free energy change, the reaction quotient, and the equilibrium constant.

LUCKY_DIMON [66]

Explanation:

Reaction quotient is defined as the ratio of the concentration of the products and reactants of a reaction at any point of time with respect to some unit. It is represented by the symbol Q.

The ratio of the concentration of products and reactants of a reaction in equilibrium with respect to some unit is said to be equilibrium constant expression. It is represented by the symbol K.

The relationship between Gibbs free energy change and reaction quotient of the reaction is:

$\Delta G=\Delta G^o+RT ln Q$ ......(1)

where,

$\Delta G$ = Gibbs free energy change

$\Delta G^o$ = Standard Gibbs free energy change

R = Gas constant

T = Temperature

At equilibrium, the free energy change of the reaction becomes 0 and standard Gibbs free energy change can be related to the equilibrium constant by the equation:

$\Delta G^o=-RT ln Q$ ...(2)

4 0

3 years ago

Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

Given:

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
Charge on the second charged particle, $q_2=3.80\ nC=3.80\times 10^{-9}\ C.$

Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

Using these values,

$\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.$

The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

The net electric field at the origin due to both the charges is given by

$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0

3 years ago

1 point

Tju [1.3M]

Answer:

Option A nuclear

Explanation:

The rate of electricity production in nuclear power plant is much higher as compared to the rate of electricity generation in gas, wind and solar power plants.

Thus, in case where large amount of electricity is to be produced in a short period then one must rely on nuclear power plants.

Therefore, option A is correct

7 0

3 years ago