All systems have what

How does the time difference between the first p-wave and the first s-wave relate to the distance to the epicenter?

Lady bird [3.3K]

S-waves do, the seismograph will detect P-waves arriving first, and S-waves will follow. The time difference, as recorded on a clock, between when the P-waves and S-waves ... earthquake waves speed up with increasing distance, and the lag time graph

6 0

3 years ago

(TCO 4) A signal consists of only two sinusoids, one of 65 Hz and one of 95 Hz. This signal is sampled at a rate of 245 Hz. Find

storchak [24]

Answer:

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

Explanation:

Given :

Frequencies of the sinusoids,

$f_{m_1}= 65 \ Hz$ , and

$f_{m_2}= 95 \ Hz$

Sampling rate $f_s = \ 245 \ Hz$

The positive frequencies at the output of the sampling system are :

$f_{o_1}=\pm f_{m_1} \pm nf_s, f_{o_2}=\pm f_{m_2} \pm nf_s$

When n = 0,

$f_{o_1}= f_{m_1} = 65 \ Hz,\ \ f_{o_2}= f_{m_2} = 95 \ Hz$

when n = 1,

$f_{o_1}=\pm f_{m_1} \pm f_s, \ \ f_{o_2}=\pm f_{m_2} \pm f_s$

$f_{o_1}= \pm 65 \pm 245,\ \ f_{o_2}=\pm 95 \pm 245$

$f_{o_1}= 180 \ Hz, 310 \ Hz,\ \ f_{o_2}= 150 \ Hz,340 \ Hz$

When n = 2,

$f_{o_1}= \pm 65 \pm 2(245),\ \ f_{o_2}=\pm 95 \pm 2(245)$

$f_{o_1}= 555 \ Hz, 425 \ Hz,\ \ f_{o_2}= 395 \ Hz,585 \ Hz$

Therefore, the first six positive frequencies present in the replicated spectrum are :

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

8 0

3 years ago

Prove that the volume of a cube that moves with a velocity vin

liq [111]

Answer:

Proof is given below

Explanation:

The length contraction is given by Δx = Δx' *√(1 - v² / c²)

where Δx' is the proper length and is measured in the frame where the object is at rest

Since the y' and z' axes are perpendicular to the direction of motion there is no contraction

So if you let V0 = Δy' * Δz' *Δx'

and V = Δy * Δz * Δx = Δy'* Δz' * Δx

Then

V = V0 * √(1 - v² / c²)

6 0

4 years ago

In a double-slit interference experiment, the slit separation is 2.41 μm, the light wavelength is 512 nm, and the separation bet

Elina [12.6K]

Answer:

Using equation 2dsinФ=n*λ

given d=2.41*10^-6m

λ=512*10^-12m

θ=52.64 degrees

7 0

4 years ago

Read 2 more answers

A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii. If the three of them are released simultaneou

Roman55 [17]

Answer:

The solid sphere will reach the bottom first.

Explanation:

In order to develop this problem and give it a correct solution, it is necessary to collect the concepts related to energy conservation. To apply this concept, we first highlight the importance of conserving energy so we will match the final and initial energies. Once this value has been obtained, we will concentrate on finding the speed, and solving what is related to the Inertia.

In this way we know that,

$\Delta KE = - \Delta PE$

$KE_t + KE_r = mgh$

We know as well that the lineal and angular energy are given by,

$KE_r = \frac{1}{2}I\omega^2$

And the tangential kinetic energy as

$KE_t = \frac{1}{2} mv^2$

Where $\omega = \frac{v}{R}$

Replacing

$\frac{1}{2}mv^2 + \frac{1}{2}I\frac{v}{R} = mgh$

Re-arrange for v,

$v=\sqrt{\frac{2mgh}{m+I/R^2}}$

We have here three different objects: solid cylinder, hollow pipe and solid sphere. We need the moment inertia of this objects and replace in the previous equation found, then,

For hollow pipe:

$I_{hp}=mR^2$