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3 years ago

How are longitudinal and transverse waves different?

Physics

1 answer:

Natalija [7]3 years ago

3 0

Answer:

A longitudinal wave is a wave where the movement of the medium is in the same direction as the wave. On the other hand, a transverse wave is a wave where the movement of the medium is at a right angle to the wave direction.

Explanation:i got this right on a quiz so i know its right

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A substance that is made of atoms of more than one type bound together is called a

BigorU [14]

The substance that is made of atoms of more than one type bound together is called an atomic bond or a multi bond

6 0

3 years ago

Assume a beam of light hits the boundary separating medium 1, with index of refraction n1 and medium 2, with index of refraction

Reptile [31]

Answer:

option C

Explanation:

Given,

Refractive index of medium 1 = n₁

Refractive index of medium 2 = n₂

For total internal reflection to take place light should move from denser medium to the rarer medium.

Here Total internal reflection take place at the boundary of medium 1 and medium 2 so, the refractive index of medium 1 is more than medium 2

n₁ > n₂

The correct answer is option C

3 0

3 years ago

Three identical 6.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up

Westkost [7]

Answer:

24 N

Explanation:

$m$ = mass of the cube = $6.0 kg$

Consider the three cubes together as one.

$M$ = mass of the three cubes together = $3 m = 3 (6.0) = 18 kg$

$a$ = acceleration of the combination = 2 ms⁻²

$F$ = Force applied on the combination

Using Newton's second law

$F = ma = (18) (2) = 36 N$

$F_{L}$ = Force by the left cube on the middle cube

Consider the forces acting on left cube, from the force diagram, we have

$F - F_{L} = ma \\36 - F_{L} = (6) (2)\\F_{L} = 24 N$

4 0

2 years ago

An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.

Olegator [25]

Hi there!

We can use Biot-Savart's Law for a moving particle:
$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }$

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }$

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }$

$B = \boxed{7.07 *10^{-10} T}$

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

$B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}$

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

$\boxed{B = 0 T}$

3 0

2 years ago

there are 1.6 km in a mile. the distance between two cities is 248 miles. How many kilometers apart are the two cities?

podryga [215]

1.6 X 248=399.117
hope this helps

3 0

2 years ago