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2 years ago

What produced the helium now present in the sun’s atmosphere? In Jupiter’s atmosphere? In the sun’s core?

Physics

1 answer:

Paul [167]2 years ago

7 0

Answer:

The helium contained in the Sun's atmosphere was produced by fusion of hydrogen to helium in the Sun's core. Since planets including the sun where all formed at the same time, Jupiter's atmosphere also contains Helium.

Explanation:

1) the helium in the suns atmosphere from the original helium and any helium the sun has produced from the nuclear fusion of hydrogen.

2) the helium present in Jupiter's atmosphere came from the helium in the matter that produced our solar system.

3) the helium found in the suns core comes only from the nuclear fusion of hydrogen

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The 2.50 kg cube in the figure has edge lengths d = 6.50 cm and is mounted on an axle

kozerog [31]

Answer:

0.191 s

Explanation:

The distance from the center of the cube to the upper corner is r = d/√2.

When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ. The new vertical distance from the center to the corner is r cos θ.

Sum of the torques:

∑τ = Iα

Fr cos θ = Iα

(k r sin θ) r cos θ = Iα

kr² sin θ cos θ = Iα

k (d²/2) sin θ cos θ = Iα

For a cube rotating about its center, I = ⅙ md².

k (d²/2) sin θ cos θ = ⅙ md² α

3k sin θ cos θ = mα

3/2 k sin(2θ) = mα

For small values of θ, sin θ ≈ θ.

3/2 k (2θ) = mα

α = (3k/m) θ

d²θ/dt² = (3k/m) θ

For this differential equation, the coefficient is the square of the angular frequency, ω².

ω² = 3k/m

ω = √(3k/m)

The period is:

T = 2π / ω

T = 2π √(m/(3k))

Given m = 2.50 kg and k = 900 N/m:

T = 2π √(2.50 kg / (3 × 900 N/m))

T = 0.191 s

The period is 0.191 seconds.

7 0

2 years ago

calculate earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the s

IceJOKER [234]

Answer:

Hello your question is incomplete below is the complete question

Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m * 1609.344 m

= 149.668 * 10^8 m

next we will express the distance between the earth and the sun

$r = \frac{a(1-E^2)}{1+Ecos\beta }$ --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

$\beta$ = 90°

input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

$v^2 = \frac{4\pi^2 }{r_{c} }$ ------ (2)

Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

( $V^2 = (\frac{4\pi^{2} }{149.626*10^9})$

therefore : V = 1.624* 10^-5 m/s

4 0

3 years ago

which form of energy does a battery-powered flashlight receive as an input? chemical A) potential energy B) radiant energy C) so

geniusboy [140]

I believe that the best answer among the choices provided by the question is the second choice ,<span>B) radiant energy

</span>
Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.

5 0

3 years ago

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If a moving car speeds up until it is going twice as fast, how much kinetic energy doe s it have compared with its initial kinet

dezoksy [38]

The kinetic energy of an object of mass m and velocity v is given by
$K= \frac{1}{2} mv^2$

Let's call $v_i$ the initial speed of the car, so that its initial kinetic energy is
$K_i = \frac{1}{2} mv_i^2$
where m is the mass of the car.

The problem says that the car speeds up until its velocity is twice the original one, so
$v_f = 2 v_i$
and by using the new velocity we can calculate the final kinetic energy of the car
$K_f = \frac{1}{2} mv_f^2 = \frac{1}{2}m (2 v_i)^2 = 4 ( \frac{1}{2} mv_i^2)=4 K_i$
so, if the velocity of the car is doubled, the new kinetic energy is 4 times the initial kinetic energy.

6 0

2 years ago

PLEASE HELP!!!! :D When you look at the Sun through a filtered telescope, the visible portion of the Sun appears blotchy.

olya-2409 [2.1K]

<span>When an individual looks through a filtered telescope in which he or she observes the sun, the portion where it appears blotchy is likely to be called the sunspots while the layer of the sun where it shows where it occurs is called the photosphere.</span>

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3 years ago

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