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2 years ago

A long wave with a period of about 15 minutes will travel across the oceans at a speed of approximately _______.

Physics

1 answer:

Temka [501]2 years ago

7 0

The speed of the wave is mathematically given as

v=2266.66m/s

A long wave with a period of about 15 minutes will travel across the oceans at a speed of approximately v=2266.66m/s

<h3>Speed of the wave</h3>

Question Parameters:

A long wave with a period of about 15 minutes

Generally the equation for the Wave velocity is mathematically given as

v=\lambda * frequency

Where

f=1/t

Therefore

v=\lambda * frequency

v=\lambda * 1/t

Therefore, with wavelenght of the ocean as 34km

v=34*1000*1/15

v=2266.66m/s

For more information on Speed

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Pand S waves are both

leva [86]

Answer:

3.body

Explanation:

Both P and S waves are both body waves. Body waves are waves that travels through the earth.

P and S waves are elastic seismic waves that moves within the earth. Love and Rayleigh waves are surface waves that travels through the surface.

P waves is also known as primary waves
S waves are known secondary waves.

These waves are the first set of waves to arrive a seismic station before the surface waves are picked up.

3 0

3 years ago

At the very end of Wagner's series of operas The Ring of Nibelung, Brunnhilde takes the golden ring form the finger of the dead

Blababa [14]

Answer:

a) 404 m² b) apparent height = 7.5 m

Explanation:

This question is about refraction and total internal refraction.

Here I will take refractive index of air and water

$n_{air}=1\\ n_{water}=1.33=4/3$

Now let's look at the diagram I have attached here

At some angle A, the light from the ring (yellow point) under water will be totally internally refracted (B = 90°), which means that rays of light (yellow arrow) that make large enough angle A will not be able to escape from the water. Since we assumed that the ring is a point, there will be a critical cone of angle A with the ring at its apex which traces a circle of radius R on the surface of water, which, beyond this radius, no light could escape.

According to snell's law

$\frac{sin(B)}{sin(A)} = \frac{n_{water}}{n_{air}} = 4/3$

At critical angle B = 90°

$\frac{3)}{4}sin(B) = [tex]\frac{3}{4} sin(90^\circ ) = 0.75 = sin(A)$

Therefore

$A = 48.6^\circ$

With this, we can find the radius of the circle (refer to my diagram)

$h* tan (A) = R\\R =11.3 m$

And with that we can find the area

$A = \pi R^2=404\ m^2$

Additional Problem

For apparent depth from above, we can think that, since we are accustomed to seeing light at the speed of c in air, our brain interpret light from <em>any</em> source to be traveling at c. This causes light that originated under water, which has the speed of

$v_{water} = \frac{c}{n_{water}} = 0.75c$

to appear as if it has traveled with the same duration as light with speed c

In order for this to happen our brain perceive shortened length which is the apparent depth.

To put it in mathematical term

$t_{travel}=\frac{h_{apparent}}{v_{water}} =\frac{h}{c}$

So we get apparent depth

$h_{apparent}=0.75h = 7.5\ m$

4 0

3 years ago

A rotating object has an angular acceleration of α = 0 rad/s2. Which one or more of the following three statements is consistent

Murrr4er [49]

Answer:

A,B and C

Explanation:

Statement A

At all times, angular velocity is $\omega = 0\,{\rm{rad/s}$

Angular acceleration is the rate of change in angular velocity with respect to time.

Angular velocity and angular acceleration are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

Which when re-arranged becomes

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

There’s no change in angular velocity anytime when the angular velocity is $\omega = 0\,{\rm{rad/s}}$

The equation can be modified as follows:

$\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero hence statement A is valid.

Statement B

Angular acceleration is the rate of change in angular velocity with respect to time.

Angular velocity and angular acceleration are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

Which when re-arranged becomes

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

There’s no change in angular velocity anytime when the angular velocity is $\omega = 10\,{\rm{rad/s}}$ .The final and initial velocities remain the same.

The equation can be modified as follows:

$\begin{array}{c}\\\alpha = \frac{{10\,{\rm{rad/s}} - 10\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero and statement B is valid

Statement C

Angular velocity is defined as the change in the angular position with respect to time.

Angular velocity and angular displacement are related by

$\theta = \omega t$

Which can also be modified as:

${\theta _{\rm{f}}} - {\theta _{\rm{i}}}$

Note that the final position is ${\theta _{\rm{f}}}$ and initial position is ${\theta _{\rm{i}}}$

Modifying the equation to find the angular velocity we obtain

$\omega = \frac{{{\theta _{\rm{f}}} - {\theta _{\rm{i}}}}}{t}$

When the angular displacement has the same value at all times, the equation becomes

$\begin{array}{c}\\\omega = \frac{{{\theta _{\rm{i}}} - {\theta _{\rm{i}}}}}{t}\\\\ = 0\\\end{array}$

The angular velocity becomes zero.

Angular acceleration and angular velocity are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

The expression above can be rearranged as follows:

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

At all times, the angular velocity is $\omega = 0\,{\rm{rad/s}}$ hence initial and final velocities remain the same

We obtain

$\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero and statement C is valid.

Therefore, statements A,B and C are consistent .

4 0

4 years ago

Explain the difference between displacement and distance

alisha [4.7K]

<span>Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.</span>

6 0

4 years ago

An electron is travelling in a straight line with a kinetic energy K = 1.60 x 10^-17J. What are (a) the magnitude and (b) the di

faust18 [17]

Answer:

(a) 1000 N/C

Explanation:

Kinetic energy of electron, K = 1.6 x 10^-17 J

distance, d = 10 cm = 0.1 m

Let the potential difference is V and the electric field is E.

(a) The relation between the kinetic energy and the potential difference is

K = e V

V = K / e

Where, e be the electronic charge = 1.6 x 10^-19 C

V = $\frac{1.6\times 10^{-17}}{1.6\times 10^{-19}}$

V = 100 V

The relation between the electric field and the potential difference is given by

V = E x d

100 = E x 0.1

E = 1000 N/C

(b) The force acting on the electron, F = q E

where q be the charge on electron

So, F = -e x E

It means the direction of electric field and the force are both opposite to each other.

The direction of electric field and the force on electron is shown in the diagram.

5 0

4 years ago