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2 years ago

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In science, Bob learns that the energy of a wave is directly proportional to the square of the waves amplitude. If the energy of

a wave is 4 J, what should he predict as the amplitude of the wave?

1 answer:

KIM [24]2 years ago

4 0

The energy of a wave is directly proportional to the square of the waves amplitude. Therefore, E = A² where A is the amplitude. This therefore means when the amplitude of a wave is doubled the energy will be quadrupled, when the amplitude is tripled the energy increases by a nine fold and so on.
Thus, in this case if the energy is 4J, then the amplitude will be √4 = 2 .

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Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

Strike441 [17]

Answer:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$

Explanation:

The electric field equation of a electromagnetic wave is given by:

$E=E_{max}(kx-\omega t)$ (1)

E(max) is the maximun value of E, it means the amplitude of the wave.
k is the wave number
ω is the angular frequency

We know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).

By definition:

$k=\frac{2\pi}{\lambda}$

$k=8.98*10^{6} [rad/m]$

And the relation between λ and f is:

$c=\lambda f$

$f=\frac{c}{\lambda}$

$f=\frac{3*10^{8}}{700*10^{-9}}$

$f=4.28*10^{14}$

The angular frequency equation is:

$\omega=2\pi f$

$\omega=2\pi*4.28*10^{14}$

$\omega=2.69*10^{15} [rad/s]$

Therefore, the E equation, suing (1), will be:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$ (2)

For the magnetic field we have the next equation:

$B=B_{max}(kx-\omega t)$ (3)

It is the same as E. Here we just need to find B(max).

We can use this equation:

$E_{max}=cB_{max}$

$B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}$

$B_{max}=1.17*10^{-8}T$

Putting this in (3), finally we will have:

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$ (4)

I hope it helps you!

8 0

3 years ago

Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect

Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a) 0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b) r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 $\pi$ $r^{2}$ = $\frac{Q1}{E_{0} }$

So,

Rearranging the above equation to get Electric field, we will get:

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$

Multiply and divide by $r1^{2}$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ x $\frac{r1^{2} }{r1^{2} }$

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

c) r > r2 :

Electric Field = ?

E x 4 $\pi$ $r^{2}$ = $\frac{Q1 + Q2}{E_{0} }$

Rearranging the above equation for E:

E = $\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

As we know from above, that:

$\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Then, Similarly,

$\frac{Q2}{E_{0} . 4 \pi. r^{2} }$ = (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

So,

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

Replacing the above equations to get E:

E = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ ) + (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Now, for

d) Under what conditions, E = 0, for r > r2?

For r > r2, E =0 if

σ1 x $r1^{2}$ = - σ2 x $r2^{2}$

4 0

3 years ago

An archer wants to hit a target that is dropped from a tower. At the sound of a horn, the archer is to shoot an arrow; at the sa

bija089 [108]

It has to be D because the arrow will drop as it moves, if it were a gun, you'd lead the target so fire below it, but due to it being an arrow, you aim high not low. Also, they didnt specify how fast anything is, so you'd probably miss if you actually did it.

6 0

2 years ago

Read 2 more answers

Why is the escape speed for a spacecraft independent of the spacecraft's mass?

scoray [572]

In order to escape the gravitational pull of our planet, any object must have an escape velocity of 7 km/s or more, anything lower than that will be slowed down by the pull of gravity, and will eventually returned to the surface of our planet. It is independent of mass, any lighter or heavier object must attain the required escaped velocity to reach space.

7 0

3 years ago

Consider f(x) = -4x2 + 24x + 3. Determine whether the function has a maximum or minimum value. Then find the

murzikaleks [220]

Answer:

The function has a maximum in $x=3$

The maximum is:

$f(3) = 39$

Explanation:

Find the first derivative of the function for the inflection point, then equal to zero and solve for x

$f(x)' = -4*2x + 24=0$

$-4*2x + 24=0$

$8x=24$

$x=3$

Now find the second derivative of the function and evaluate at x = 3.

If $f (3) ''< 0$ the function has a maximum

If $f (3) '' >0$ the function has a minimum

$f(x)''= 8$

Note that:

$f(3)''= -8$

the function has a maximum in $x=3$

The maximum is:

$f(3)=-4(3)^2+24(3) + 3\\\\f(3) = 39$

4 0

3 years ago