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polet [3.4K]
2 years ago
12

Two coherent sources of radio waves, A and B, are 5.00 metersapart. Each source emits waves with wavelength 6.00 meters.Consider

points along the line connecting the two sources.
a) At what distance from source A is there constructiveinterference between points A and B?
b) At what distances from source A is there destructiveinterference between points A and B?
Physics
1 answer:
alexgriva [62]2 years ago
3 0

Answer

given,

distance between A and B is equal to 5 m

let x be the distance of loud speaker from the source A

For constructive interference, path difference = mλ

x - (5 - x ) = 0           , for m = 0

2 x - 5 = 0

2 x = 5

  x = 2.5 cm

For destructive interference

|x - (5 - x )| = (2 m + 1)\dfrac{\lambda}{2}

m = 0

|2x - 5| = \dfrac{\lambda}{2}

|2x - 5| = \dfrac{6}{2}

\pm(2x - 5) = 3

x = 4 , 1 cm

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A 1,103 kg car traveling at 18 m/s to the south collides with a 4,919 kg truck that is initially at rest at a stoplight. The car
Zigmanuir [339]

ANSWER

3.30\text{ m/s}

EXPLANATION

Parameters given:

Mass of car, mc = 1103 kg

Mass of truck, mt = 4919 kg

Initial velocity of car, uc = 18 m/s

Inital velocity of truck = 0 m/s

To solve this problem, we have to apply the law of conservation of momentum, which states that the total momentum of a system is constant.

This implies that:

m_cu_c+m_tu_t=m_cv_c+m_tv_t

Since the car and the truck stick together after the collision, they will have the same final velocity.

Hence:

m_cu_c+m_tu_t=(m_c+m_t)v_{}_{}

Substitute the given values and solve for v (final velocity):

\begin{gathered} (1103\cdot18)+(4919\cdot0)=(1103+4919)v \\ \Rightarrow19854=6022v \\ \Rightarrow v=\frac{19854}{6022} \\ v=3.30\text{ m/s} \end{gathered}

That is the final velocity of the two-vehicle mass.

8 0
10 months ago
Two particles are fixed to an x axis: particle 1 of charge q1 = 2.78 × 10-8 c at x = 15.0 cm and particle 2 of charge q2 = -3.24
Oksi-84 [34.3K]
Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
E=\frac{1}{4\pi\varepsilon_0}\frac{q}{(x-x')^2}
Now At xp we have:
\frac{1}{4\pi\varepsilon_0}\frac{q_1}{(x_p-x_1)^2}-\frac{1}{4\pi\varepsilon_0}\frac{3.29q_1}{(x_p-x_2)^2}=0
\implies (x_p-x_1)^2=\frac{(x_p-x_2)^2}{3.29}\\
\implies(1-\frac{1}{3.29})x_p^2+2(\frac{x_2}{3.29}-x_1)x_p+x_1^2-\frac{x_2^2}{3.29}=0
Which is a second order equation, using the quadratic formula to solve for xp would give us:
xp=\frac{-(\frac{x_2}{3.29}-x_1)-\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}
or
xp=\frac{-(\frac{x_2}{3.29}-x_1)+\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.

Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.

6 0
3 years ago
At which boundary does subduction stop occurring, resulting in a mountain range?
NARA [144]
Subduction occurs at convergent tectonic plate boundaries, resulting in high rates of volcanism, earthquakes and mountain building.
6 0
2 years ago
Read 2 more answers
¿Cuál es la velocidad del sonido en el aíre a -25°C?
Svetradugi [14.3K]

La velocidad del sonido en el aire (a una temperatura de 20 ºC) es de 343 m/s. La ecuación creada por Newton y posteriormente modificada por Laplace que permite obtener la velocidad del sonido en el aire teniendo en cuenta la variable de la temperatura es "331+(0,6 x Temperatura)".

               

8 0
3 years ago
Three equal point charges, each with charge 1.45 μCμC , are placed at the vertices of an equilateral triangle whose sides are of
LUCKY_DIMON [66]

Answer:

U = 80.91 J

Explanation:

In order to calculate the electric potential energy between the three charges you use the following formula:

U=k\frac{q_1q_2}{r_{1,2}}                  (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q1: q2 charge

r1,2: distance between charges 1 and 2.

For the three charges you have:

U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}           (2)

You use the fact that q1=q2=q3=q and that the distance between charges are equal. Then, in the equation (2) you have:

q = 1.45μC = 1.45*10^-6C

r = 0.700mm = 0.700*10^-3m

U_T=3k\frac{q^2}{r}=3(8.98*10^9Nm^2/C^2)\frac{(1.45*10^{-6}C)}{0.700*10^{-3}m}\\\\U_T=80.91J

The electric potential energy between the three charges is 80.91 J

7 0
3 years ago
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