All categories

3 years ago

The mountains most commonly found at divergent plate boundaries are

Physics

1 answer:

dybincka [34]3 years ago

8 0

fault-block mountains. hope this helps

You might be interested in

could anyone determinate the period knowing that it performs 4000 vibrations in 0.5 minutes, Sorry my english is bad

dexar [7]

Answer:

f = 4000 / 30 sec = 133.3 vibrations/sec

P = 1 / f = .0075 sec period of 1 vibration

4 0

2 years ago

Can someone please help me

Mars2501 [29]

The correct answer is letter b.

To find the answer follow the following steps.
1. 6524.96 x .25 = X

2. 1631.24 = X

This works for all of the given answers to find the correct answer.

5 0

3 years ago

Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average o

Musya8 [376]

Explanation:

It is given that,

A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

Radius of Mercury's orbit, $r=5.79\times 10^{10}\ m$

Radius of discovered planet, $R=\dfrac{2}{3}r$

$R=\dfrac{2}{3}\times 5.79\times 10^{10}\ m=3.86\times 10^{10}\ m$

Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :

$T^2\propto R^3$

$T^2=\dfrac{4\pi^2R^3}{GM}$

$T^2=\dfrac{4\pi^2\times (3.86\times 10^{10})^3}{6.67\times 10^{-11}\times 1.99\times 10^{30}}$

$T=\sqrt{1.71\times 10^{13}}$

T = 4135214.625 s

T = 47.86 days

So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.

6 0

3 years ago

Light enters a substance from air at 30 degrees to thenormal.

Gwar [14]

Answer:

Explanation:

The angle of incidence and refraction are both measured from the normal

angle of incidence = 30°

angle of refraction = 23°

refractive index(n) = sini / sinr

n = sin30°/sin23°

n = 1.27965

refractive index (n) = 1/sinC

where C is the critical angle.

sinC= 1/n

C =arcsin (1/n)

C =arcsin (1/1.27965)

C = 51.39°

5 0

4 years ago

What is the weight on mass

Gre4nikov [31]

Answer:

6.39×10^23 kg is the weight on mass

3 0

2 years ago