All categories

2 years ago

Two long parallel wires are placed side-by-side on a horizontal table. if the wires carry current in opposite directions,

Physics

1 answer:

STatiana [176]2 years ago

3 0

$\vec F = I (\vec L \times \vec B)$

if currents go in opposite directions, wires repel

You might be interested in

What is the period (time) that corresponds to oscillatory motion with a frequency of 23.4 Hz when the speed of light is 3.0×10^8

OLEGan [10]

"Period" and "frequency" of waves and oscillations are simple reciprocals. They're not related to the wavelength or speed.

Period = 1 / frequency

Period = 1 / 23.4 Hz

<em>Period = 0.043 second</em>

8 0

3 years ago

The ability of atoms to attract electrons in chemical bonds is called

Marina CMI [18]

It is electronegativity the property of an atom which increases with its tendency to attract the electrons of an bond

4 0

3 years ago

Explain what is the difference of physical and chemical weathering

shusha [124]

The difference is that physical weathering is a process that weathers rock without a chemical reaction or change. Chemical weathering changes the identity of rocks and it involves a chemical reaction or change.

5 0

2 years ago

Four charges of magnitude +q are placed at the corners of a square whose sides have a length d. What is the magnitude of the tot

guapka [62]

Answer:

$F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}$

Explanation:

Since all the four charges are equidistant from the position of Q

so here we can assume this charge distribution to be uniform same as that of a ring

so here electric field due to ring on its axis is given as

$E = \frac{k(4q)x}{(x^2 + R^2)^{1.5}}$

here we have

x = b

and the radius of equivalent ring is given as the distance of each corner to the center of square

$R = \frac{d}{\sqrt2}$

now we have

$E = \frac{4kq b}{(b^2 + \frac{d^2}{2})^{1.5}}$

so the force on the charge is given as

$F = QE$

$F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}$

3 0

3 years ago

Consider two children sitting on a merry-go-round, with one closer to the outer edge and one closer to the center. show answer N

dolphi86 [110]

Answer:

They both have the same angular speed.

Explanation:

The mathematical formula for angular speed is:

$w=\frac{2\pi}{T}$

where $w$ is angular speed, $2\pi$ is a constant, and $T$ is the period (the time it takes the marry-go-round to complete a lap).

What we can see from the formula is that, since the $2\pi$ does not change its value, the angular speed depends only on the period T.

In this case for both the children closer to the outher edge and for the children closer to the center, the time to complete a lap is the same, because the time does not depend on where they are sitting in the marry go round. This means that the period for both is the same.

Thus, since the period for both is the same, the angular speed given by

$w=\frac{2\pi}{T}$ will also be the same

4 0

3 years ago