All categories

2 years ago

How do p-wave and S-wave differ?

Physics

1 answer:

oksian1 [2.3K]2 years ago

3 0

Answer:

P wave has pee lol just kidding one of them transmit more sound or intrections than the other

Explanation:

You might be interested in

A bowling ball is traveling at 3 m/s. It starts to roll up a ramp. How high above the ground will the ball be when it stops roll

Zepler [3.9K]

Answer:

FREE FREE FREE FREEE FREEEEEEEEEEE FREEEEEEEEEEEEEEEE FREEEEEEEEEEEEEEEEEE

Explanation:

FREE FREE FREE FREEE FREEEEEEEEEEE FREEEEEEEEEEEEEEEE FREEEEEEEEEEEEEEEEEE

8 0

2 years ago

Read 2 more answers

What are the subzones for each main zone?

Makovka662 [10]

Answer:

The oceanic zone is subdivided into the epipelagic, mesopelagic, and bathypelagic zones on the basis of amount of light that reaches different depths. The mesopelagic (disphotic) zone, where only small amounts of light penetrate, lies below the Epipelagic zone.

Explanation:

5 0

2 years ago

Light of wavelength 600 nm passes though two slits separated by 0.22mm and is observed on a screen 1.1m behind the slits. The lo

Ne4ueva [31]

Answer:

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

Explanation:

We have to take into account the expression for the position of the fringes

$dsen\theta=m\lambda\\y=\frac{m\lambda D}{d}$

where m is the number of the maximum, d is the separation of the slits, D is the distance to the screen.

(a) By replacing we obtain

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

(b) more information is required to solve this point. Please complete the information.

HOPE THIS HELPS!

4 0

3 years ago

Read 2 more answers

What type of charge does an proton have?

Whitepunk [10]

A proton has a positive (1+) charge

4 0

3 years ago

Read 2 more answers

a coil has n turns enclosing an area of a. in a physics laboratory experiment, the coil is rotated during the time interval δt f

Mrac [35]

The correct answer for the question that is being presented above is this one:

Phi = BAsin(theta)
1. Phi(i) = BA 
2. Phi(f) = 0 
3. EMF = N(phi(i)-phi(f))/deltat

Here are the follow-up questions:

1. What is the total magnitude Phi_initial of the magnetic flux through the coil before it is rotated? 
2. What is the magnitude Phi_final of the total magnetic flux through the coil after it is rotated? 
3. What is the magnitude of the average emf induced in the coil?

8 0

2 years ago