If a star were located exactly at each celestial pole, the corrected altitude of the star would equal _________

A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice

Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of $\sqrt{3} v_{e}$

<u>Explanation:</u>

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , $v_{i}=2 v_{e}$ . Here the radius is R, so find the escape velocity as follows,

$\frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0$

$\frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}$

$v_{e}^{2}=\frac{2 G M}{R}$

$v_{e}=\sqrt{\frac{2 G M}{R}}$

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as, $U_{i}=-\frac{G M m}{R}$

R tend to infinity when far away from the planet, so $v_{f}=0$

Then, kinetic energy at initial would be,

$k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}$

Similarly, kinetic energy at final would be,

$k_{f}=\frac{1}{2} m v_{f}^{2}$

Here, $v_{f}=\text { final velocity }$

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

$U_{i}+k_{i}=k_{f}+v_{f}$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}$

'm' and $\frac{1}{2}$ as common on both sides, so gets cancelled, we get as

$4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}$

We know, $v_{e}=\sqrt{\frac{2 G M}{R}}$ , it can be wriiten as $\left(v_{e}\right)^{2}=\frac{2 G M}{R}$ , we get

$4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}$

$v_{f}^{2}=3\left(v_{e}\right)^{2}$

Taking squares out, we get,

$v_{f}=\sqrt{3} v_{e}$

4 0

3 years ago

How does refraction allow thin convex lenses to work?

melomori [17]

Answer:

https://www.khanacademy.org/science/physics/geometric-optics/lenses/v/convex-lenses

Explanation:

Here is a link to a video to tell you about this.

7 0

3 years ago

Read 2 more answers

The bricks are homogeneous and have the same mass. Each brick is 5.00 cm

Kitty [74]

Answer:

(5/3,-5/6) or (1.67, -0.83)

Explanation:

the center if there are 4 bricks is the point 0,0 or intersection of the two arrow.

if we remove one on the top left, with the center of the mass in (-5,2.5), the the new coordinates of the center of mass can be found like this

x = A1 x1 - A2 x2 / (A1 - A2)

x = new coordinate of the system

A1 is the Area of 4 bricks = 20 x 10 = 200 cm²

A2 is the Area of 1 brick we remove, the top left = 10 x 5 = 50 cm²

x1 = the x coordinate of the center of mass when there are 4 bricks, x = 0

x2 = the x coordinate of the center of mass of the remove brick, x = -5

x = (200 . 0) - (50 . (-5))/(200 - 50)

x = - (-250) / 150

x = 250/150 = 5/3 = 1.67 cm

we can find the value y of the new coordinate, the same way we find x

y = A1 y1 - A2 y2 / (A1 - A2)

y = (200 . 0) - (50 . 2.5) / (200-50)

y = - 125/150

y = -5/6 = -0.83 cm

4 0

3 years ago

A ball is thrown straight up with a velocity of 16 m/s; what will be its velocity 2.0s after being released?

harkovskaia [24]

The answer for the question is 8m/s

4 0

3 years ago

the fundamental frequency for the 3rd chord on a five-string guitar is 240 Hz. what frequency would produce the 10th harmonic?

Mamont248 [21]

The frequency of the 10th harmonic is 800 Hz

Explanation:

The frequency of the nth-harmonic for the standing waves in a string is given by the equation

$f_n = n f_1$

where

$f_1$ is the fundamental frequency of the string

In this problem, we are given the frequency of the 3rd harmonic:

$f_3 = 240 Hz$

Which can be rewritten in terms of the fundamental frequency

$f_3 = 3 f_1$

So we find $f_1$ :

$f_1 = \frac{f_3}{3}=\frac{240}{3}=80 Hz$

Now that we have the fundamental frequency, we can find the frequency of the 10th harmonic, with n = 10 :

$f_{10} = 10f_1 = (10)(80)=800 Hz$

Learn more about waves:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

6 0

4 years ago

If a star were located exactly at each celestial pole, the corrected altitude of the star would equal __________.