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3 years ago

Describe how Newton’s third law applies to the box and the table.

Physics

2 answers:

Pepsi [2]3 years ago

5 0

Answer:

An object at rest remains at rest, an object in motion remains in motion at a constant velocity unless affected by external forces. So the box will remain at rest since the acceleration is zero

Snowcat [4.5K]3 years ago

3 0

Answer:

the box has vertical force on the table

Explanation:

There can be no single isolated force, for every action there must be a force of reaction of ingual magnitude and direction, but in the opposite direction

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Ira Lisetskai [31]

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3 years ago

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

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ludmilkaskok [199]

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Guys answer with a clear explanation and plzz don't spam.

timama [110]

Answer:

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Explanation:

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