All categories

3 years ago

Is there 1,000 more milky way in the galaxy???If it is true is there life also?

2 answers:

5 0

The second part of the question can not be fully answered because it is not proven that there is or isn’t others life forms in this universe. personallyi think that there is but whatva

Yuri [45]3 years ago

3 0

The "Milky Way" is the name we call OUR Galaxy ... the one we're in. There are several billion OTHER galaxies besides the one we're in.

There are SO MANY stars in the Milky Way, and SO MANY other galaxies, and SO MANY stars in every other Galaxy, and SO MANY planets around SO MANY stars, that most scientists believe that it's a sure bet that there's life in other places, or that there used to be, and it's a sure bet that aliens exist in other places, or that they used to.

But we've never seen ANY evidence of ANY life in ANY other place except Earth.

Yet.

You might be interested in

if an average cloud has a density of 0.5 g/m3 and has a volume of 1,000,000,000 m3,what is the weight of an average cloud

sveta [45]

If you have (1 x 10⁹) cubic meters of volume, and 1/2 gram of mass
in each cubic meter, then you have

(1/2 x 10⁹) grams = (5 x 10⁸) grams = (5 x 10⁵) kilograms of mass .

On Earth, that mass weighs 4,900,000 newtons .

(about 1,102,300 pounds)

(about 551 tons)

3 0

4 years ago

Is cold fusion science or pseudoscience

kondor19780726 [428]

The experiments that claimed to demonstrate cold fusion were found
to have been faulty by others who reviewed them. Also, nobody else
was able to reproduce the finding in other laboratories. In the world
of Science, this pretty much says that the initial claims were unfounded.

5 0

3 years ago

Help ;-;

nasty-shy [4]

Answer:

all qn 1,2,3 have same answer ,. Yes,. hope it helps

3 0

3 years ago

A parallel-plate capacitor has a voltage of 391 v applied across its plates, then the voltage source is removed. what is the vol

andrezito [222]

When the capacitor is connected to the voltage, a charge Q is stored on its plates. Calling $C_0$ the capacitance of the capacitor in air, the charge Q, the capacitance $C_0$ and the voltage ( $V_0=391 V$ ) are related by

$C_0 =\frac{Q}{V_0}$ (1)

when the source is disconnected the charge Q remains on the capacitor.

When the space between the plates is filled with mica, the capacitance of the capacitor increases by a factor 5.4 (the permittivity of the mica compared to that of the air):

$C=k C_0 = 5.4 C_0$

this is the new capacitance. Since the charge Q on the plates remains the same, by using eq. (1) we can find the new voltage across the capacitor:

$V=\frac{Q}{C}=\frac{Q}{5.4 C_0}$

And since $Q=C_0 V_0$ , substituting into the previous equation, we find:

$V=\frac{C_0 V_0}{5.4 C_0}=\frac{V_0}{5.4}=\frac{391 V}{5.4}=72.4 V$

7 0

3 years ago

For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

5 0

3 years ago