How does a message travel across the gap at a synapse?

zvonat [6]

Before Pluto was discovered, it was predicted. Astronomers had observed that massive objects can affect the orbits of its neighbors, and, after seeing deviations in the orbits of Uranus and Neptune, assumed something substantial existed beyond their orbits.
When Pluto was spotted, it was thought to be the predicted object and was identified as a ninth planet.
A few decades later, astronomers started discovering more and more objects around other stars and didn’t know whether to call them planets or not. There appeared to be a need to define what a planet means, and that led to what some people consider Pluto’s demotion to a dwarf planet.
The International Astronomical Union decided that full-sized planets must orbit the sun, have a round shape, and have cleared their orbits of other objects. Pluto fulfills the first two criteria, but not the third.
It still goes around the sun, it’s round enough, it’s got moons, and behaves like a planet, but the idea is that Pluto did not form the same way as the rest of the planets. Pluto’s orbit is both eccentric and inclined more than the rest of the planets by about 17 degrees. That’s suggests something is different about this object.
This debate about whether to call it a planet or not is silly, because it doesn’t matter to Pluto what you call it. It is an interesting object, goes around the sun, and shows geology and an atmosphere.
There’s a tendency to define objects based on what they are now, but nothing is constant in the universe. There are some issues with the nomenclature, and a definition today may not apply to the same object tomorrow.

7 0

3 years ago

Read 2 more answers

6. One minute after takeoff, a rocket carrying the space shuttle into outer space reaches a speed of 447 m/s.

Sidana [21]

Answer:

acceleration of the rocket is given as

$a = 7.45 m/s^2$

Explanation:

As we know that rocket starts from rest and then reach to final speed of 447 m/s after t = 1 min

so we have

$v_i = 0$

$v_f = 447 m/s$

$t = 1 min = 60 s$

so we have

$a = \frac{v_f - v_i}{t}$

$a = \frac{447 - 0}{60}$

$a = 7.45 m/s^2$

7 0

3 years ago

An airplanes lands with a velocity of +75 m/s and comes to a stop in 20 seconds. What is the acceleration of the airplane while

erma4kov [3.2K]

Answer:

The acceleration of the car, a = -3.75 m/s²

Explanation:

Given data,

The initial velocity of the airplane, u = 75 m/s

The final velocity of the plane, v = 0 m/s

The time period of motion, t = 20 s

Using the I equations of motion

v = u + at

a = (v - u) / t

= (0 - 75) / 20

= -3.75 m/s²

The negative sign indicates that the plane is decelerating

Hence, the acceleration of the car, a = -3.75 m/s²

7 0

3 years ago

A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

$f'=(\frac{v+v_r}{v+v_s})f$

where

f is the original frequency

f is the apparent frequency

$v$ is the velocity of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

$v_s = v_A = -28.7 m/s$ (surface A is the source, which is moving towards the receiver)

$v_r = +62.2 m/s$ (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

$f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz$

(b) 0.232 m

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

$\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m$

(c) 1481.2 Hz

Again, we can use the same formula

$f'=(\frac{v+v_r}{v+v_s})f$

In the source frame (= on surface A), we have

$v_s = v_B = -62.2 m/s$ (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

$v_r = +28.7 m/s$ (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

$f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz$

(d) 0.225 m

The wavelength of the wave is given by

$\lambda=\frac{v}{f}$

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

8 0

3 years ago

20 POINT QUESTION

Valentin [98]

Label A: sublimation
Label B: condensation
Label C: melting

8 0

3 years ago

Read 2 more answers