Ask question

Ask question

All categories

3 years ago

9

A lifeguard on a beach observes that waves have a speed of 7 m/s and a distance of 4.6m between wave crests. what is the period

of the wave motion?

1 answer:

Pepsi [2]3 years ago

4 0

Answer:

0.657 seconds

Explanation:

speed of wave= wavelength / time period

so

time period= wavelength / speed

= 4.6/7

=0.657 sec

You might be interested in

a sound pulse emitted underwater reflects off a school of fish and is detected at the same place 0.01 s later. how far away are

andrew-mc [135]

In fresh water sound waves travel at 1497m/s at 25 degrees, I'll assume that's the characteristics of the water.

If it's 0.01s then you need to divide the speed by 100 to get the, 14.97, however it gets there and back in that time so you need to halve it.
<u>7.485m</u>

5 0

3 years ago

Picture question please answer do in 5 mins

alexandr402 [8]

Answer:

Can't see anything, please share clearly

4 0

3 years ago

As a wave travels through a medium, it displaces particles in a direction parallel to the motion of the wave. We can conclude th

postnew [5]

We can conclude that it is a longitudinal wave because the wave is traveling through a medium displacing particles<span>
</span>

8 0

3 years ago

Read 2 more answers

Please help :)<br><br><br> Microwaves are transmitted by ....... ?

Dima020 [189]

Microwaves are transmitted by Radio Waves.

4 0

3 years ago

A disk of mass M and radius R rotates at angular velocity ω0. Another disk of mass M and radius r is dropped on top of the rotat

AleksandrR [38]

Answer:

$\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}$

Explanation:

As we know that there is no external torque on the system of two disc

then the angular momentum of the system will remains conserved

So we will have

$L_i = L_f$

now we have

$L_i = (\frac{1}{2}MR^2)\omega_o$

also we have

$L_f = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega$

now from above equation we have

$(\frac{1}{2}MR^2)\omega_o = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega$

now we have

$\omega = \frac{MR^2\omega_o}{(MR^2 + Mr^2)}$

$\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}$

6 0

3 years ago