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2 years ago

In a compression wave, particles in the medium move

Physics

1 answer:

nikdorinn [45]2 years ago

8 0

in the same direction as the wave

Explanation:

In a compression wave, the particles in the medium moves in the same direction as the wave source.

A wave is generally defined as a disturbance that transmits energy.

There are two types of waves based on the direction through which they are propagated.
Transverse waves are directed perpendicularly in the direction of propagation.
Examples are electromagnetic waves.
Longitudinal waves are parallel to their source. Examples are sound waves, p-waves.
They are made up of series of rarefaction and compression.

learn more:

Waves brainly.com/question/3183125

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Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small

Vaselesa [24]

Answer:

The final kinetic energy of the Helium nucleus (alpha particle) after been scattered through an angle of 120° is

8.00 x 10-13J

Explanation:

In Rutherford Scattering experiment, the collision of the helium nucleus with the gold nucleus is an ELASTIC COLLISION. This means that the kinetic energy is conserved ( The same before and after the collision).

Thus, the final kinetic energy of the helium nucleus is the same as initial kinetic energy (8.00 x 10^-13Joules)

Although, the kinetic energy is converted to potential energy in Coulomb's law equation.

That is,

1/2(mv^2) = (K* q1q2)/r

Where m is the mass of helium nucleus, v is its colliding velocity, k is electrostatic constant, q1 is the charge on helium nucleus, q2 is the charge on gold nucleus, r is impact parameter

6 0

2 years ago

Assume a satellite shines an unpolarized light on a telescope. The intensity of the light as it reaches the telescope is 1.1*10-

n200080 [17]

Answer:

$4.125\times 10^{-11}\ W/m^2$

Explanation:

$I_0$ = Intensity of unpolarized light = $1.1\times 10^{-10}\ W/m^2$

$\theta$ = Angle of the filter = $30^{\circ}$

Intensity of light is given by

$I=\dfrac{I_0}{2}cos^2\theta\\\Rightarrow I=\dfrac{1.1\times 10^{-10}}{2}cos^230\\\Rightarrow I=4.125\times 10^{-11}\ W/m^2$

The intensity of light detected by the camera is $4.125\times 10^{-11}\ W/m^2$

7 0

3 years ago

A transverse wave on a string is described by the following wave function.y = (0.090 m) sin (px/11 + 4pt)(a) Determine the trans

alukav5142 [94]

Explanation:

(a) It is known that equation for transverse wave is given as follows.

y = $(0.09 m)sin(\pi \frac{x}{11} + 4 \pi t)$

Now, we will compare above equation with the standard form of transeverse wave equation,

y = $A sin(kx + \omega t)$

where, A is the amplitude = 0.09 m

k is the wave vector = $\frac{\pi}{11}$

$\omega$ is the angular frequency = $4\pi$

x is displacement = 1.40 m

t is the time = 0.16 s

Now, we will differentiate the equation with respect to t as follows.

The speed of the wave will be:

v(t) = $\frac{dy}{dt}$

v(t) = $A \omega cos(kx + \omega t)$

v(t) = $(0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)$

v(t) = -0.84 m/s

The acceleration of the particle in the location is

a(t) = $\frac{dv}{dt}$

a(t) = $-A \omega 2sin(kx + \omega t)$

a(t) = $-(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)$

a(t) = -9.49 $m/s^{2}$

Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 $m/s^{2}$ .

(b) Wavelength of the wave is given as follows.

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

$\lambda = (frac{2\pi}{\frac{\pi}{11})
$

$\lambda$ = 22 m

The period of the wave is

T = $\frac{2 \pi}{\omega}$

T = $\frac{2 \pi}{4 \pi}$

= 0.5 sec

Now, we will calculate the speed of propagation of wave as follows.

v = $\frac{\lambda}{T}$

= $\frac{22 m}{0.5 s}$

= 44 m/s

therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.

7 0

3 years ago

11. A cyclist accelerates from 0 m/s to 10 m/s in 3 seconds. What is his acceleration ? Is this acceleration higher than that of

Marat540 [252]

$a = \frac{v - u}{t}$

v = final velocity

u = initial velocity

t = time taken

the acceleration of the cyclist is

$\frac{10 - 0}{3} = 3.333333....$

approximately 3.33 m/s^2

the acceleration of the car is

$\frac{40 - 0 }{8} = 5.0$

5.0 m/s^2

$5.0 > 3.33 \\ so \: the \: answer \: is \: no$

6 0

2 years ago

A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of the plank hanging over the ri

lawyer [7]

Answer:

x = 0.6034 m

Explanation:

Given

L = 5 m

Wplank = 225 N

Wman = 522 N

d = 1.1 m

x = ?

We have to take sum of torques about the right support point. If the board is just about to tip, the normal force from the left support will be going to zero. So the only torques come from the weight of the plank and the weight of the man.

∑τ = 0 ⇒ τ₁ + τ₂ = 0

Torque come from the weight of the plank = τ₁

Torque come from the weight of the man = τ₂

⇒ τ₁ = + (5 - 1.1)*(225/5)*((5 - 1.1)/2) - (1.1)*(225/5)*((1.1)/2) = 315 N-m (counterclockwise)

⇒ τ₂ = Wman*x = 522 N*x (clockwise)

then

τ₁ + τ₂ = (315 N-m) + (- 522 N*x) = 0

⇒ x = 0.6034 m

7 0

2 years ago