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3 years ago

8

Coulomb’s Law relates which of the following?

1 answer:

Verdich [7]3 years ago

8 0

There's nothing on the list you provided that's related to Coulomb's Law.

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when the heat is added to the material, which factors are important in order to determine how much the material will rise in tem

posledela

The heat Q transferred to cause a temperature change depends on the magnitude of the temperature change, the mass of the system, and the substance and phase involved.

Explanation:

https://courses.lumenlearning.com/physics/chapter/14-2-temperature-change-and-heat-capacity/

5 0

3 years ago

The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi

Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

$\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$

Where,

B= Magnetic Field

l = length

$\mu_0$ = Vacuum permeability

$\epsilon_0$ = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

$B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}$

Recall that the speed of light is equivalent to

$c^2 = \frac{1}{\mu_0 \epsilon_0}$

Then replacing,

$B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}$

$B = \frac{r}{2C^2} \frac{dE}{dt}$

Our values are given as

$dE = 2150N/C$

$dt = 5s$

$C = 3*10^8m/s$

$D = 0.440m \rightarrow r = 0.220m$

Replacing we have,

$B = \frac{r}{2C^2} \frac{dE}{dt}$

$B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}$

$B =5.25*10^{-16}T$

Therefore the magnetic field around this circular area is $B =5.25*10^{-16}T$

3 0

3 years ago

Can you help quick in science please

nikdorinn [45]

Uhhhh...you should have paid attention in class, just saying...

4 0

3 years ago

The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m

Simora [160]

Answer:

The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

$F = k \frac{q_1 q_2}{d^2}$

where k is Coulomb's constant, $q_1$ and $q_2$ are the charges and d is the distance between the charges.

Working a little the equation, we can take:

$d^2 = k \frac{q_1 q_2}{F}$

$d = \sqrt{ k \frac{q_1 q_2}{F}}$

And this equation will give us the distance between the charges. Taking the values of the problem

$k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00$

(the force has a minus sign, as its attractive)

$d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}$

$d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}$

$d = \sqrt{ 28,462,500 \ m^2}}$

$d = 5,335.026 m$

And this is the distance between the charges.

3 0

3 years ago

In which situation will the lowest resistance occur?

insens350 [35]

Answer:

thick wire and cold temperatures

8 0

3 years ago

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