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Sergio039 [100]
3 years ago
8

1. Which of the following statements about mechanical waves is true?

Physics
1 answer:
uranmaximum [27]3 years ago
5 0
1) <span>a. mechanical waves require a medium to travel through
In fact, mechanical waves cannot travel trough empty space, but they always require a medium for their propagation. (on the contrary, electromagnetic waves can also travel through empty space)

2) d</span><span>. gamma rays
In fact, gamma rays are the radiations with shortest wavelength on the electromagnetic spectrum (range in the picometer). Correspondingly, this is the radiation with highest energy, because the energy of the radiation is inversely proportional to its wavelength, according to:
</span>E=h \frac{c}{\lambda}
where h is the planck constant, c the speed of light and \lambda the wavelength.

3) <span>d. Infrared have lower frequency than x-rays
In fact, infrared radiation has frequency in the range of the THz (terahertz, </span>10^{12} Hz), while x-rays have frequency in the range of hundreds of PHz (Petahertz, 10^{15}Hz).

4) <span>d. parallel
In longitudinal waves, the oscillations of the wave (such as the motion of the particles in a compression wave) occur in the direction parallel to the direction of propagation of the wave itself.

5) </span><span>c. gases
Sound travels slowest in gases. In fact, the speed of sound is proportional to the density of the medium: the more dense the medium is, the faster is the sound in that medium, and vice-versa. Since gases are less dense than liquids and solids, sound travels slowest in gases. The reason is that sounds propagates through the oscillations of the molecules of the medium; if the medium is less dense (such as in gases), it takes more time to transmit the sound through the medium itself.</span>
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Two charges are located in the x-y plane. If q1=-4.55 nC and is located at x=0.00 m, y=0.680 m and the second charge has magnitu
Elden [556K]

Answer:

Ex= -23.8 N/C  Ey = 74.3 N/C

Explanation:

As the  electric force is linear, and the electric field, by definition, is just this electric force per unit charge, we can use the superposition principle to get the electric field produced by both charges at any point, as the other charge were not present.

So, we can first the field due to q1, as follows:

Due  to q₁ is negative, and located on the y axis, the field due to this charge will be pointing upward, (like the attractive force between q1 and the positive test charge that gives the direction to the field), as follows:

E₁ = k*(4.55 nC) / r₁²

If we choose the upward direction as the positive one (+y), we can find both components of E₁ as follows:

E₁ₓ = 0   E₁y = 9*10⁹*4.55*10⁻⁹ / (0.68)²m² = 88.6 N/C (1)

For the field due to q₂, we need first to get the distance along a straight line, between q2 and the origin.

It will be just the pythagorean distance between the points located at the coordinates (1.00, 0.600 m) and (0,0), as follows:

r₂² = 1²m² + (0.6)²m² = 1.36 m²

The magnitude of the electric field due to  q2 can be found as follows:

E₂ = k*q₂ / r₂² = 9*10⁹*(4.2)*10⁹ / 1.36 = 27.8 N/C (2)

Due to q2 is positive, the force on the positive test charge will be repulsive, so E₂ will point away from q2, to the left and downwards.

In order to get the x and y components of E₂, we need to get the projections of E₂ over the x and y axis, as follows:

E₂ₓ = E₂* cosθ, E₂y = E₂*sin θ

the  cosine of  θ, is just, by definition, the opposite  of x/r₂:

⇒ cos θ =- (1.00 m / √1.36 m²) =- (1.00 / 1.17) = -0.855

By the same token, sin θ can be obtained as follows:

sin θ = - (0.6 m / 1.17 m) = -0.513

⇒E₂ₓ = 27.8 N/C * (-0.855) = -23.8 N/C (pointing to the left) (3)

⇒E₂y = 27.8 N/C * (-0.513) = -14.3 N/C (pointing downward) (4)

The total x and y components due to both charges are just the sum of the components of Ex and Ey:

Ex = E₁ₓ + E₂ₓ = 0 + (-23.8 N/C) = -23.8 N/C

From (1) and (4), we can get Ey:

Ey = E₁y + E₂y =  88.6 N/C + (-14.3 N/C) =74.3 N/C

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