All categories

3 years ago

Why do astronomers believe that triton is a captured moon?

1 answer:

6 0

Answer: Accoding to research "Triton is unique among all the large moons in the solar system because it orbits Neptune in a direction opposite to the planet's rotation (a "retrograde" orbit). It is unlikely to have formed in this configuration and was probably captured from elsewhere."

Explanation:

You might be interested in

How does wavelength change depending on whether the pipe is open or closed?

il63 [147K]

Answer:

* in a tube with both ends open

λ = 2L / n n = 1, 2, 3,…

* tube with one end open and the other closed.

λ = 4L / n n = (2n ’+ 1 )

Explanation:

When a sound wave enters a pipe, a resonance process occurs, whereby only some wavelengths can occur.

* in a tube with both ends open

in this case there are maximums at each end, so if the length of the had is L

λ = 2L 1st harmonic

λ = 2 L / 2 2nd harmonic

λ = 2L / 3

λ = 2L / n n = 1, 2, 3,…

* In the case of a tube with one end open and the other closed.

At the open end there is a belly and at the closed end a node

λ = 4L 1st harmonic

λ = 4L / 3 3rd harmonic

λ = 4L / 5

λ = 4L / n n = 1, 3, 5,… odd

n = (2n ’+ 1 ) where n’ are all integers

4 0

3 years ago

A car accelerates from rest at 3m/s2 along a straight road how far has the car traveled after 4s?

murzikaleks [220]

Answer:

24 m

Explanation:

$x = 1/2at^{2} + v_{0}t + x_{0}\\a = 3 m/s^{2}\\v_{0} = 0$

7 0

3 years ago

A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how

GalinKa [24]

Answer: 473.640 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=48.1 m/s$ is the cannonball's initial velocity

$\theta=0$ because we are told the cannonball is shot horizontally

$t$ is the time since the cannonball is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=1.5m$ is the initial height of the cannonball

$y=0$ is the final height of the cannonball (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it $X_{max}$ and this occurs when $y=0$ .

So, firstly we will find the time with (2):

$0=1.5 m+48. 1 m/s sin(0\°) t-(4.9 m/s^{2})t^2$ (3)

Rearranging the equation:

$0=-(4.9 m/s^{2})t^2+48. 1 m/s sin(0\°) t+1.5 m$ (4)

$-(4.9 m/s^{2})t^2+(48. 1 m/s) t+1.5 m=0$ (5)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula: