All categories

3 years ago

How does a large star create the elements? at what element does the star stop making new elements?

1 answer:

4 0

They are fused in the core of the star due to great pressures and temperatures. They are made all the way through iron. At that point the star dies. If it is a really large star it will become a supernova when it dies, creating all of the elements beyond iron as well, but only in its death. No star can create anything beyond iron in its life cycle

You might be interested in

A wooden plaque is in the shape of an ellipse with height 30 centimeters and width 22 centimeters. Find an equation for the elli

Volgvan

Answer:

Explanation:

height of Ellipse $=30 cm$

i.e. $2 a=30$

Width of Ellipse $=22 cm$

i.e. $2 b=22$

Equation of a vertical Ellipse is

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

$a=15 ,b =11$

$\frac{x^2}{11^2}+\frac{y^2}{15^2}=1$

at $y=4 cm$

$\frac{x^2}{11^2}+\frac{4^2}{15^2}=1$

$x=\frac{11}{15}\times \sqrt{15^2-4^2}$

$x=10.6 cm$

5 0

2 years ago

A car slows down from 80 km/h to 60 km/h in 2 seconds.

Andreyy89

Answer:

please mark brainlist

Explanation:

Initial speed u=80 km/h=80×185=22.22 m/s

Final speed v=60 km/h=60×185=16.67 m/s

Using v=u+at

Or 16.67=22.22+α×5

⟹ a=−1.1 m/s2

6 0

3 years ago

Using 0.500 g of nichrome, you are asked to fabricate a wire with uniform cross-section. The resistance of the wire is 0.673 Ω.

mojhsa [17]

Explanation:

Given that,

Mass of Nichrome, m = 0.5 g

The resistance of the wire, R = 0.673 ohms

Resistivity of the nichrome wire, $\rho=10^{-6}\ \Omega -m$

Density, $d=8.31\times 10^3\ kg/m^3$

(A) The length of the wire is given by using the definition of resistance as :

Volume,

$V=A\times l\\\\A=\dfrac{V}{l}\\\\Since, V=\dfrac{m}{d}\\\\V=\dfrac{m}{d}\\\\V=\dfrac{0.5\times 10^{-3}}{8.31\times 10^3}\\\\V=6.01\times 10^{-8}\ m^3$

Area,

$A=\dfrac{V}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{l}$ ....(1)

$R=\rho \dfrac{l}{A}\\\\l=\dfrac{RA}{\rho}\\\\l=\dfrac{0.673\times 6.01\times 10^{-8}}{l\times 10^{-6}}\\\\l=0.201\ m$

(b) Equation (1) becomes :

$A=\dfrac{6.01\times 10^{-8}}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{0.201}\\\\\pi r^2=3\times 10^{-7}\\\\r=\sqrt{\dfrac{3\times 10^{-7}}{\pi}} \\\\r=3.09\times 10^{-4}\ m$

Hence, this is the required solution.

5 0

3 years ago

suppose that 273 g of one of the substances listed above displaces 26 mL of water. What is the substance?

guajiro [1.7K]

<span>The unknown substance is silver. I don't see a list of available substances, but let's see if there's something reasonable available that will match. First, let's calculate the density of the unknown substance. Density is mass per volume, so 273 g / 26 mL = 10.5 g/mL Looking up a list of elements sorted by density, I see the following: 10.07 Actinium 10.22 Molybdenum 10.5 Silver 11.35 Lead And silver at 10.5 g/ml is a very nice match for the unknown substances' density of 10.5 g/ml.</span>

6 0

3 years ago

Read 2 more answers

You drop a rock down a well and hear a splash 3 s later. As Charlie Brown would say, the well is 3 seconds deep. How fast is the

vampirchik [111]

Since it was dropped, it should be the speed of gravity which is 9.8 meters/second

5 0

1 year ago