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2 years ago

Seismic waves do not travel along the Earth’s surface. Please select the best answer from the choices provided T F.

Physics

1 answer:

zysi [14]2 years ago

7 0

Seismic waves do not travel along the Earth’s surface. The statement is false.

<h3>what are seismic waves?</h3>

The seismic waves can be defined as the waves full of energy that travel through the Earth's layer, these waves are the result of earthquake.

They can further be divided into two types body wave and surface waves. The body waves travel in the interior of the Earth and the surface waves travel at the surface of the Earth.

The seismic waves can have an amplitude of few centimeters in large earthquake. They travel slower as compared to the body waves.

Hence Seismic waves do not travel along the Earth’s surface. The statement is false.

To know more about Seismic waves follow

brainly.com/question/6835759

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Why car speed is not considered as car velocity?

grigory [225]

Answer:

Cause its scalar quantity

Explanation:

since speed does not take directions into consideration, it is considered to be a scalar quantity. On the other hand, the velocity of an object does not take into account direction, thus making it a vector quantity.

4 0

4 years ago

A sled of mass m is being pulled horizontally by a constant horizontal force of magnitude F. The coefficient of kinetic friction

rusak2 [61]

I'll bite:

-- Since the sled's mass is 'm', its weight is 'mg'.

-- Since the coefficient of kinetic friction is μk, the force acting opposite to the direction it's sliding is    (μk) times (mg) .

-- If the pulling force is constant 'F', then the horizontal forces on the sled
are 'F' forward and (μk · mg) backwards.

-- The net force on the sled is (F - μk·mg).
(I regret the visual appearance that's beginning to emerge,
but let's forge onward.)

-- The sled's horizontal acceleration is (net force) / (mass) = (F - μk·mg) / m.
This could be simplified, but let's not just yet.

-- Starting from rest, the sled moves a distance 's' during time 't'.
We know that s = 1/2 a t² , and we know what 'a' is. So we can write

   s = (1/2 t²) (F - μk·mg) / m .

Now we have the distance, and the constant force.
The total work is (Force x distance), and the power is (Work / time).
Let's put it together and see how ugly it becomes. Maybe THEN
it can be simplified.

Work = (Force x distance) = F x (1/2 t²) (F - μk·mg) / m

Power = (Work / time) = <em>F (t/2) (F - μk·mg) / m </em>

Unless I can come up with something a lot simpler, that's the answer.

To simplify and beautify, make the partial fractions out of the
2nd parentheses:
   <em> F (t/2) (F/m - μk·m)</em>

I think that's about as far as you can go. I tried some other presentations,
and didn't find anything that's much simpler.

Five points,ehhh ?

4 0

3 years ago

Read 2 more answers

A sprinter runs 500 meters west in a straight line and then turns around and runs in the same

Kipish [7]

Answer:

200 meters

Explanation:

6 0

4 years ago

An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.

Olegator [25]

Hi there!

We can use Biot-Savart's Law for a moving particle:
$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }$

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }$

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }$

$B = \boxed{7.07 *10^{-10} T}$

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

$B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}$

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

$\boxed{B = 0 T}$

3 0

2 years ago

What rapid changes to earths surface are caused by shifting plates

PtichkaEL [24]

I am not sure if this is the answer you are looking for but a earthquake occurs when the plates shift.
Hope this helped.

4 0

3 years ago