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3 years ago

Choose the letter of the best description of the nature of light:

Physics

1 answer:

Andreas93 [3]3 years ago

6 0

C. Light sometimes behaves like waves and at other times like particles.

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6. Velocity is a ____________. a) vector b) value c) arrow d) none of these

Olenka [21]

<span>the speed of something in a given direction. so i think none of these</span>

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3 years ago

Read 2 more answers

The formula K2S indicates that

jekas [21]

There are two atoms of potassium bonded to one atom of sulfur.

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3 years ago

What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo

Phantasy [73]

Answer:

The magnetic field will be $\large{\dfrac{1.4 \times 10^{-4}}{d}} T$ , '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field ( $\large{\overrightarrow{B}}$ ) at a distance ' $r$ ' due to a current carrying conductor carrying current ' $I$ ' is given by

$\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}$

where ' $\overrightarrow{dl}$ ' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current ' $I$ ', at a distance ' $d$ ' is given by

$\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}$

According to the figure if ' $I_{t}$ ' be the current carried by the top wire, ' $I_{b}$ ' be the current carried by the bottom wire and ' $2d$ ' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the $\bigotimes$ symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by $\bigodot$ symbol.

Given $\large{I_{t} = 19.5 A}$ and $\large{I_{B} = 12.5 A}$

Therefore, the magnetic field ( $\large{B_{t}}$ ) at 'P' due to the top wire

$B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}$

and the magnetic field ( $\large{B_{b}}$ ) at 'P' due to the bottom wire

$B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}$

Therefore taking the value of $\mu_{0} = 4\pi \times 10^{-7}$ the net magnetic field ( $\large{B_{M}}$ ) at the midway between the wires will be

$\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T$

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3 years ago

How long does it take a 750-w coffeepot to bring to a boil 0.75 l of water initially at 11°c? assume that the part of the pot wh

Evgen [1.6K]

Just add all of them up and there is your answer I just added it but I want u to work it out to..

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3 years ago

A particle moving in the x direction is being acted upon by a net force f(x)=cx2, for some constant

makkiz [27]

The work-energy theorem states that the change in kinetic energy of the particle is equal to the work done on the particle:
$\Delta K = W$
The work done on the particle is the integral of the force on dx:
$W= \int\limits^{3L}_L {F(x)} \, dx = \int\limits^{3L}_L {cx^2} \, dx = \frac{26}{3}cL^3$
So, this corresponds to the change in kinetic energy of the particle.

6 0

3 years ago