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1 year ago

Why is simple harmonic motion a fundamental for understanding physics

1 answer:

6 0

The simple harmonic motion not only describes circular movements, also describes waves motion. This makes it fundamental for physics, make us understand waves like sound or light which explains an enormus part of our environment

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Which statement best describes electrons

Brums [2.3K]

The statement that best describes electrons is that t<span>hey are negative subatomic particles and are found surrounding the nucleus.</span>

7 0

3 years ago

Which of the three types of equations follows the law of conservation of mass?

ArbitrLikvidat [17]

Answer:

BE MORE DESCRIPTIVE

Explanation:

8 0

3 years ago

A thick, spherical shell made of solid metal has an inner radius a = 0.18 m and an outer radius b = 0.46 m, and is initially unc

KonstantinChe [14]

(a) $E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

We can solve the different part of the problem by using Gauss theorem.

Considering a Gaussian spherical surface with radius r<a (inside the shell), we can write:

$E(r) \cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

where q is the charge contained in the spherical surface, so

$q=5.00 C$

Solving for E(r), we find the expression of the field for r<a:

$E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

(b) 0

The electric field strength in the region a < r < b is zero. This is due to the fact that the charge +q placed at the center of the shell induces an opposite charge -q on the inner surface of the shell (r=a), while the outer surface of the shell (r=b) will acquire a net charge of +q.

So, if we use Gauss theorem for the region a < r < b, we get

$E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}$

however, the charge q' contained in the Gaussian sphere of radius r is now the sum of the charge at the centre (+q) and the charge induced on the inner surface of the shell (-q), so

q' = + q - q = 0

And so we find

E(r) = 0

(c) $E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

We can use again Gauss theorem:

$E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}$ (1)

where this time r > b (outside the shell), so the gaussian surface this time contained:

- the charge +q at the centre

- the inner surface, with a charge of -q

- the outer surface, with a charge of +q

So the net charge is

q' = +q -q +q = +q

And so solving (1) we find

$E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

which is identical to the expression of the field inside the shell.

(d) $-12.3 C/m^2$

We said that at r = a, a charge of -q is induced. The induced charge density will be

$\sigma_a = \frac{-q}{4\pi a^2}$

where $4 \pi a^2$ is the area of the inner surface of the shell. Substituting

q = 5.00 C

a = 0.18 m

We find the induced charge density:

$\sigma_a = \frac{-5.00 C}{4\pi (0.18 m)^2}=-12.3 C/m^2$

(e) $-1.9 C/m^2$

We said that at r = b, a charge of +q is induced. The induced charge density will be

$\sigma_b = \frac{+q}{4\pi b^2}$

where $4 \pi b^2$ is the area of the outer surface of the shell. Substituting

q = 5.00 C

b = 0.46 m

We find the induced charge density:

$\sigma_b = \frac{+5.00 C}{4\pi (0.46 m)^2}=-1.9 C/m^2$

3 0

3 years ago

The type of function that describes the amplitude of damped oscillatory motion is _______. The type of function that describes t

Salsk061 [2.6K]

Answer:

exponential

Explanation:

type of function that describes the amplitude of damped oscillatory motion is exponential because as we know that here function is

y = A × $e^{\frac{-bt}{2m}}$ × cos(ωt + ∅ ) ..................................... ( 1 )

here function A × $e^{\frac{-bt}{2m}}$ is amplitude

as per equation ( 1 )it is exponential

so that we can say that amplitude of damped oscillatory motion is exponential

8 0

3 years ago

g As a proton moves in the direction the electric field lines A) it is moving from low potential to high potential and gaining e

Yuki888 [10]

Answer:

Option D => it is moving from high potential to low potential and losing electric potential energy.

Explanation:

Consider a big circle, within the circle we have force, F. That force, F is known as the Electric Field and inside the region or field or space, charged particle or object will be able to exerts force on the other objects.

Electric Field can be represented mathematically by using the formula below; E = kQ/r^2.

So, let us answer the question with what we have considered above. It is worthy of note to know that electric Field moves from a region of higher potential to a region of lower potential. So, any option that says this is correct.

But, there is only one problem and that is the fact that the question asked us about the direction of the movement of proton. Since, proton is s a positive charge, it is going to lose electric potential energy. So, Option D is correct.

3 0

3 years ago