All categories

3 years ago

Could an average star, such as our sun, become a neutron star? Explain your answer.

1 answer:

5 0

<span>No. Neutron stars are the remnants of very large stars that have supernova'd. Anything below 1.44 solar masses becomes a dwarf, anything above 5 solar masses becomes a black hole. Everything in between becomes a neutron star (or quark star, but it's not proven).</span>

You might be interested in

The first confirmed detections of extrasolar planets occurred in ____________. The first confirmed detections of extrasolar plan

nirvana33 [79]

Answer:

1992 (Early 1990s)

Explanation:

First of all, i would like to define an extrasolar planet as a planet that orbits a start that is not our own.

The first confirmed detections of extrasolar planets occured in the early 1990s (specifically 1992, some say 1995). The name of the first extrasolar planet is widely believed to be called Dimidium or 51 Pegasi b.

Extrasolar were searched by monitoring stars for slight dimming that might occur as unseen planets pass in front of them.

4 0

3 years ago

Two objects are dropped from rest from the same height. Object A falls through a distance Da and during a time t, and object B f

stiv31 [10]

Answer:

Da=(1/4)Db

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

When s = Da, t = t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times t^2\\\Rightarrow Da=\frac{1}{2}at^2$

When s = Db, t = 2t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times (2t)^2\\\Rightarrow Db=\frac{1}{2}a4t^2$

Dividing the two equations

$\frac{Da}{Db}=\frac{\frac{1}{2}at^2}{\frac{1}{2}a4t^2}=\frac{1}{4}\\\Rightarrow \frac{Da}{Db}=\frac{1}{4}\\\Rightarrow Da=\frac{1}{4}Db$

Hence, Da=(1/4)Db

3 0

3 years ago

In odd nuclei, what determines the final spin of the nucleus?

Katen [24]

Answer:

In odd nuclei, the left out proton or neutron will contribute to the spin of the nucleus.

Explanation:

The meaning of odd nuclei is atomic mass is odd.

A=odd number.

A=Z+n

Here, Z is proton either it will odd or n will odd which is neutron.

Now according to the shell model the left out proton or neutron will contribute to the spin and parity.

For example,

Take the case of isotope of nitrogen-15.

Here Z is 7, and n is 8 will not contribute in spin.

Now, for Z=7.

$1S^{2} _{\frac{1}{2} }, 1P^{4} _{\frac{3}{2} }, 1P^{1} _{\frac{1}{2} }$

Here,

$j=\frac{1}{2}$

and, L=1.

Fort parity,

$(-1)^{L}$

Put the value of L.

Parity will be -1.

Now, spin will be

$S=(\frac{1}{2} )^{-1}$ .

7 0

3 years ago

Light with an intensity of 1 kW/m2 falls normally on a surface and is completely absorbed. The radiation pressure is

kobusy [5.1K]

Answer:

The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

Explanation:

Given;

intensity of light, I = 1 kW/m²

The radiation pressure of light is given as;

$Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}$

I kW = 1000 J/s

The energy flux density = 1000 J/m².s

The speed of light = 3 x 10⁸ m/s

Thus, the radiation pressure of the light is calculated as;

$Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa$

Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

6 0

3 years ago

Physical quantity are measurable quantities

Vilka [71]

Answer:

yeah physical quantities are the quantities which can be meaured

7 0

3 years ago