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2 years ago

This diagram shows the process that powers stars. This process is called?

Physics

2 answers:

Elanso [62]2 years ago

5 0

Answer: The correct answer is Option C.

Explanation:

A star is defined as a fixed point in the night sky which illuminates and is very large. the primary element present in stars is Hydrogen and it continuously changes to helium by a process known as Nuclear fusion reaction. This process releases a huge amount of energy.

Nuclear fusion reaction is defined as the combination of lighter nuclei to form a heavy nuclei. In Stars, hydrogen atoms combine to form a helium atom.

The reactions which takes place in stars follows a series of equations:

$_1^1\textrm{H}+_1^1\textrm{H}\rightarrow _1^2\textrm{H}+_{+1}^0\textrm{e}+\text{energy}\\\\_1^2\textrm{H}+_1^1\textrm{H}\rightarrow _2^3\textrm{He}+\text{energy}\\\\_2^3\textrm{He}+_1^1\textrm{H}\rightarrow _2^4\textrm{He}+_{+1}^0\textrm{e}+\text{energy}$

Overall reaction for the above series of reactions is given by:

$4_1^1\textrm{H}\rightarrow _2^4\textrm{He}+2_{+1}^0\textrm{e}+\text{energy}$

Hence, the correct answer is Option C.

viktelen [127]2 years ago

4 0

The process that powers stars is C)fusion

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Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),

yuradex [85]

Answer:

$v=3.564\ m.s^{-1}$

$\Delta v =2.16\ m.s^{-1}$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

Using the equation of motion:

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3$

$v_J=5.94\ m.s^{-1}$

Now using the law of conservation of momentum at the bottom of the slide:

Sum of initial momentum of kids before & after collision must be equal.

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30\times 5.94+0=(30+20)v$

$v=3.564\ m.s^{-1}$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

Now we find the force along the slide due to the body weight:

$F=m_J.g.sin\theta$

$F=30\times 9.8\times \frac{1.8}{3}$

$F=176.4\ N$

Hence the net force along the slide:

$F_R=71.4\ N$

Now the acceleration of John:

$a_j=\frac{F_R}{m_J}$

$a_j=\frac{71.4}{30}$

$a_j=2.38\ m.s^{-2}$

Now the new velocity:

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2\times 2.38\times 3$

$v_J_n=3.78\ m.s^{-1}$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^{-1}$

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2 years ago

A standard 1 kilogram weight is a cylinder 41.5 mm in height and 44.0 mm in diameter. what is the density of the material?

Korvikt [17]

Answer;

=15855.40 kg/m^3

Explanation;

Volume (V) of the cylinder = pi x r^2 x h

V = 3.14 x (44/2 x 10^-3)^2 x 41.5 x 10^-3

V = 6.307 x 10^-5 m^3

By density = m/V

mass = 1 kg

density = 1/(6.307 x 10^-5) = 15855.40 kg/m^3

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