What are two ways in which all types of precipitation are alike

Physics

1 answer:

Digiron [165]3 years ago

7 0

1. they all come from the clouds 2.they are all formed by precipatation. hope this helps enjoyyyyyy!!!!!!!!!!!!!

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A 290-turn solenoid having a length of 32 cm and a diameter of 11 cm carries a current of 0.30 A. Calculate the magnitude of the

iris [78.8K]

The magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.

To find the answer, we need to know about the magnetic field inside the solenoid.

<h3>What's the expression of magnetic field inside a solenoid?</h3>

Mathematically, the expression of magnetic field inside the solenoid= μ₀×n×I

n = no. of turns per unit length and I = current through the solenoid

<h3>What's is the magnetic field inside the solenoid here?</h3>

Here, n = 290/32cm or 290/0.32 = 906

I= 0.3 A

So, Magnetic field= 4π×10^(-7)×906×0.3 = 3.4×10^(-4) T.

Thus, we can conclude that the magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.

Learn more about the magnetic field inside the solenoid here:

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6 0

1 year ago

What is the wavelength of an earthquake wave if it has a speed of 12 km/s and a frequency of 15 Hz

professor190 [17]

First change km/ s into m/s, then use the formula
Lambda = velocity/ frequency

3 0

3 years ago

#1 Not sure where to start. This is for AP Physics!

yaroslaw [1]

First,

$\rho=\dfrac mV$

where $\rho$ is density, $m$ is mass, and $V$ is volume. We can compute the volume of the roll:

$2.7\,\dfrac{\mathrm g}{\mathrm{cm}^3}=\dfrac{1275\,\mathrm g}V$

$\implies V\approx472.22\,\mathrm{cm}^3\approx4.72\,\mathrm m^3$

When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness $x$ . Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).

So we have

$V=Ax$

where $A$ is the given area, so

$4.72\,\mathrm m^3=\left(18.5\,\mathrm m^2\right)x$

$\implies x\approx0.255\,\mathrm m=255\,\mathrm{mm}$

If we're taking significant digits into account, the volume we found would have been $V=470\,\mathrm m^3$ , in turn making the thickness $x=250\,\mathrm{mm}$ .