All categories

4 years ago

Could we see a galaxy that is 20 billion light-years away? (assume that we mean a "lookback time" of 20 billion years.)

1 answer:

3 0

yes we could its just that the star is most likely dead when we see it i know we can see a star like this because even if the star dies the light is still traveling, like if you went on a drive away from your home and then your home is destroyed just as you leave, just because you house is gone does not mean you cant still travel.

You might be interested in

What is the definition of velocity? (AKS 1/DOK1)

allsm [11]

Explanation:

The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of an object's speed and direction of motion

5 0

3 years ago

Read 2 more answers

A proton moves along the x-axis in the laboratory with velocity uy = 0.6c. An observer moves with a velocity of v=0.8c along the

Dima020 [189]

Explanation:

Given that,

Velocity of the proton in lab frame $u_{x} = 0.6c$

Velocity of the observer v= 0.8c

We need to calculate the velocity of the proton with respect to the observer

Using formula of velocity

$u'=\dfrac{v-u_{x}}{1-\dfrac{u_{x}v}{c^2}}$

$u'=\dfrac{0.8c-0.6c}{1-\dfrac{0.6c\times0.8c}{c^2}}$

$u'=c(\dfrac{0.8-0.6}{1-(0.8\times0.6)})$

$u'=0.385c$

(a). We need to calculate the total energy of the proton in the lab frame

Using formula of kinetic energy

$K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u_{x}^2}{c^2}}}-1)$

Where, Proton mass energy = m₀c²

Put the value into the formula

$K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.6c)^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.6)^2}}-1)$

$K.E=234.57\ MeV$

(b). We need to calculate the kinetic energy of the proton in the observer

Using formula of kinetic energy

$K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u'^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.385)^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.385)^2}}-1)$

$K.E=78.366\ MeV$

(c). We need to calculate the momentum of the proton with respect to observer

Using formula of momentum

$P_{obs}=\dfrac{m_{0}u'^2}{\sqrt{1-\dfrac{u'^2}{c^2}}}$

We know that,

Proton mass energy = m₀c²

$m_{0}=\dfrac{938.28}{c^2}$

$P_{obs}=\dfrac{\dfrac{938.28}{c^2}\times(0.385c)^2}{\sqrt{1-\dfrac{(0.385c)^2}{c^2}}}$

$P_{obs}=\dfrac{938.28\times(0.385)^2}{\sqrt{1-0.385^2}}$

$P_{obs}=150.69\ MeV/c$

Hence, This is required solution.

3 0

3 years ago

The boiling point of water at sea level is 100 °c. at higher altitudes, the boiling point of water will be

Scilla [17]

The boiling point of water at sea level is 100 °C. At higher altitudes, the boiling point of water will be.....
a) higher, because the altitude is greater.
b) lower, because temperatures are lower.
c) the same, because water always boils at 100 °C.
d) higher, because there are fewer water molecules in the air.
==> e) lower, because the atmospheric pressure is lower.
--------------------------
Water boils at a lower temperature on top of a mountain because there is less air pressure on the molecules.
-------------------
I hope this is helpful.

8 0

3 years ago

PLEASE HELP i really need help with these

enyata [817]

Answer:

Explanation:

IDK

7 0

3 years ago

Read 2 more answers

If 0.600 mol of a nonvolatile nonelectrolyte are dissolved in 3.70 mol of water, what is the vapor pressure ph2o of the resultin

fredd [130]

To get the vapor pressure of the resulting solution, we use the Raoult’s Law:

Psolution = (χsolvent) (P°solvent)

1st: Calculate the mol fraction of the solvent. In this case, water is the solvent.

χsolvent = 3.7 mol / (3.7 mol + 0.6 mol)

χsolvent = 0.86

2nd: Calculate the vapor pressure.

Psolution = (0.86) (23.8 torr)

Psolution = 20.48 torr

4 0

3 years ago