All categories

3 years ago

In the United States, where can volcanoes be found?

1 answer:

7 0

<span>The volcanoes of USA are located along the west coast, at the subduction of the Pacific and North American tectonic plates. During the 20th century there were only two eruptions in continental USA; Lassen in 1915, and Mt St Helens in 1980. Alaska and Hawaii contain currently active volcanoes. hope this helps:)</span>

You might be interested in

Some people will realize that they are dreaming while the dream is happening this is called

algol [13]

Answer:

C. Lucid Dreaming

6 0

3 years ago

Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$

Here

$M = \text{Mass inside the Orbit of the star}$

$R = \text{Orbital radius}$

$G = \text{Universal Gravitational Constant}$

Mass inside the orbit in terms of Volume and Density is

$M =V \rho$

Where,

V = Volume

$\rho =$ Density

Now considering the volume of the star as a Sphere we have

$V = \frac{4}{3} \pi R^3$

Replacing at the previous equation we have,

$M = (\frac{4}{3}\pi R^3)\rho$

Now replacing the mass at the gravitational acceleration formula we have that

$a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho$

$a_g = \frac{4}{3} G\pi R\rho$

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

$a_c = \frac{4}{3} G\pi R\rho$

At the same time the general expression for the centripetal acceleration is

$a_c = \frac{\Theta^2}{R}$

Where $\Theta$ is the orbital velocity

Using this expression in the left hand side of the equation we have that

$\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2$

$\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}$

$\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R$

Considering the constant values we have that

$\Theta = \text{Constant} \times R$

$\Theta \propto R$

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0

3 years ago

Does this marble have potential energy?<br> 1) no<br> 2)yes

jek_recluse [69]

Answer:

yes

Explanation:

because it has the potential to move

5 0

3 years ago

The force F⃗ pulling the string is constant; therefore the magnitude of the angular acceleration α of the wheel is constant for

abruzzese [7]

Answer:

The answer is " $\boxed{\boxed{\omega = \sqrt{\frac{2fd}{kmr^2}}}}$ "

Explanation:

$\to d= r \theta \\\\ \to \theta =\frac{d}{r}\\\\\to \omega^{r} - \omega_{0}^{r} = 2 \alpha \theta\\\\\to \omega^{r} = 2 \alpha \theta - \omega_{0}^{r} \\\\\to \omega^{r} = 2 (\frac{F}{Kmr}) \frac{d}{r}\\\\\to \omega = \sqrt{\frac{2fd}{kmr^2}}$