3 years ago

What happens to the atoms in a star as it goes through its life?

Physics

2 answers:

DochEvi [55]3 years ago

8 0

A the atoms at the beginning of a starts life are the same as the end of its life

Alina [70]3 years ago

4 0

Answer:

The correct answer is going to be C

Explanation:

You might be interested in

Based on the Punnett square, what percentage of offspring would have genotype YY?

mel-nik [20]

50 c should be right

6 0

3 years ago

Explain how you would draw magnetic field lines in the vicinity of this magnet (it is a normal north and south magnet)

sammy [17]

Can someone please answer this question I'm doing the test now?

3 0

3 years ago

Which segments show changes of state that absorb heat? Check all that apply.

Nat2105 [25]

Answer:

B-C

D-E

Explanation:

Trust

3 0

3 years ago

When placed at a certain point, a

Wewaii [24]

Answer: 180

Explanation: Acellus

6 0

3 years ago

A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude

Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$ .

Now, let us call $\lambda$ the charge per unit length, then we know that

$dQ = \lambda Rd\theta$ ;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda }{R}.$

Now, we know that

$\lambda = 3.0*10^{-9}C/m,$

$k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},$

and the radius of the semicircle is

$\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };$

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9}) }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

7 0

3 years ago