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2 years ago

Last one I swear, 50 points again! plz help this is schoolwork

2 answers:

8 0

It is a stretch of the atmosphere ranging from the upper mesosphere to the lower parts of the thermosphere. It’s useful to us in radio communication.

shutvik [7]2 years ago

6 0

Answer:

Where is the ionosphere, and what makes it useful to us? The ionosphere is a stretch of the atmosphere ranging from the upper mesosphere to the lower parts of the thermosphere. It is useful to us in radio communication, as radio signals can bounce off of it to extend their range

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A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

6 0

3 years ago

Which is the electric potential energy of a charged particle divided by its charge?Electric fieldelectric field lineelectric pot

Dahasolnce [82]

<h3><u>Answer;</u></h3>

electric potential

<h3><u>Explanation;</u></h3>

Electric potential is the electric potential energy per unit charge.

Mathematically; V =PE/q

Where; PE is the electric potential energy, V is the electric potential and q is the charge.

Electric potential is more commonly known as voltage. If you know the potential at a point, and you then place a charge at that point, the potential energy associated with that charge in that potential is simply the charge multiplied by the potential.

7 0

3 years ago

Frank does 2400J of work in climbing a set of stairs. If he does the work in 6 seconds, what is his power output?

timurjin [86]

Power = Work Done / Time taken

Where Work Done is Joules, and Time is in Seconds, Power is in Watts

= 2400 J / 6 seconds

= 400 Watts

The power output is 400 Watts.

4 0

3 years ago

What does fossil fuels do?

katrin [286]

Fossil fuel are collected and turned into oils like car gas and many other products have fossil fuels in them and we don't even realize it idk oil was made out of fossils till i studied fossils in class Hope this helps:)

8 0

3 years ago

Read 2 more answers

A gun shoots a bullet with a velocity of 500 m/s. The gun is aimed horizontally and fired from a height of 1.5 m. How far does t

MrMuchimi

The bullet travels a horizontal distance of 276.5 m

The bullet is shot forward with a horizontal velocity $u_x$ . It takes a time <em>t</em> to fall a vertical distance <em>y</em> and at the same time travels a horizontal distance <em>x. </em>

The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.

The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.

Calculate the time taken for the bullet to fall through a vertical distance <em>y </em>using the equation,

$y=u_yt+\frac{1}{2} gt^2$