All categories

3 years ago

A transform boundary occurs where two tectonic plates _____.?

2 answers:

8 0

- - B. Move apart from one another.

This is because A transform boundary occurs where two tectonic plates split together and move the opposite sides from each other.

✡ Hope this helps! ✡

gladu [14]3 years ago

7 0

Where they slide over each other.

Transform boundaries are formed or occur when two plates slide past each other in a sideways motion. They do not tear or crunch into each other (but the rock in between them may be ground up) and therefore none of the spectacular features are seen such as occur in divergent and convergent boundaries.

In a transform boundary, neither plate is added to at the boundary nor destroyed. They are marked in some places by features like stream beds that have been split in half and the two halves moved in opposite directions.

You might be interested in

I only need the answer to #21

Oksanka [162]

X-axis=time
y-axis=distance (position)

slope=y-axis / x-axis=distance / time=velocity.

answer: d) slope

6 0

3 years ago

In a light wave, what properties tell you the color of light

sesenic [268]

Answer:

Answer :- In a light wave the property of wave which tells about the color of light is it's Wavelength .

Wavelength is the distance between one crest and one through , also it is the distance after which the wave repeat itself !

It's SI unit is meter !

It is scalar quantity !!

Different Wavelength of light have different color !!

• VIBGYOR

i.e, Violent , Indigo , Blue , Green , Yellow Orange, and Red along with their shades are the colors which we can see !!

• They almost range from 400nm to 700nm ( visible range of light ) !!

4 0

3 years ago

The condition required to work to be done

Dvinal [7]

Answer:

work=f(costheta)

Explanation:

work is done when a force acts on a body and displaces it on the direction of force

6 0

3 years ago

A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s

Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

Time of flight = t = 20 s
Angle of the initial velocity of projectile with the horizontal = $\theta = 30^\circ$

<u>Assume:</u>

Initial velocity of the projectile = u
R = Range of the projectile during the time of flight
H = maximum height of the projectile
D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

Initial horizontal velocity of the projectile = $u\cos \theta$
Initial horizontal velocity of the projectile = $u\sin \theta$

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

$\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s$

Hence, the initial speed of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

$\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m$

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

$R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m$

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

$\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m$

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

6 0

3 years ago

Which of the following statements cannot be supported by Kepler's laws of planetary motion? a. The distance of a planet around t

hodyreva [135]

i guess option (a)

is the right one

if it's correct

select me as brainliest

5 0

3 years ago