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2 years ago

Which force is responsible for making fusion possible in the sun?

Physics

2 answers:

GrogVix [38]2 years ago

7 0

What are the answer choices, can you tell me the please?

Veseljchak [2.6K]2 years ago

3 0

Answer: Strong Nuclear Force is responsible for making fusion possible in the sun .

Explanation:

Fusion reaction is basically the result of the interplay of two opposing forces

1. Attractive Strong Nuclear Force

2. Repulsive coulomb force

Strong nuclear force is a short range force while its magnitude is much greater than the coulomb force . When two nuclei approach each other coulomb force of repulsion between the nuclei (proton-proton repulsion) acts between the nuclei but due high speed as the nuclei come closer strong nuclear force of attraction come into play which has magnitude much more larger than the coulomb force .

The strong force grows rapidly once the nuclei are close enough, and the fusing nucleons can essentially "fall" into each other and the result is fusion and net energy produced. Thus it is the presence of Strong Nuclear Force which makes the fusion possible .

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EXPLAIN HOW ENERGY IS TRANSMITTED THROUGH A MEDIUM

jek_recluse [69]

I say it helped then because TrueType had room

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3 years ago

Read 2 more answers

Atoms are happy (they will not readily react with other elements) when they have a full outside ring of

Len [333]

Answer: TRUE

Explanation:

Atoms are happy when they will not react with other elements while having a full outside ring of electrons because this makes them to be noble.

A stable atom possesses full outside ring of electrons while unstable one does not. So, they are happy also because of stability.

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3 years ago

Cual de las escalas de temperatura es la mas antigua

IrinaVladis [17]

Answer:

the translation I got for this question is

Which of the temperature scales is the oldest?

Explanation:

and i searched for it and got this=

Fahrenheit scale

6 0

2 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago

A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. What is the amplitude of the m

Anna35 [415]

To solve this problem it is necessary to take into account the concepts of Intensity as a function of Power and the definition of magnetic field.

The intensity depending on the power is defined as

$I = \frac{P}{4\pi r^2},$

Where

P = Power

r = Radius

Replacing the values that we have,

$I = \frac{60}{(4*\pi (0.7)^2)}$

$I = 9.75 Watt/m^2$

The definition of intensity tells us that,

$I = \frac{1}{2}\frac{B_o^2 c}{\mu}$

Where,

$B_0 =$ Magnetic field

$\mu =$ Permeability constant

c = Speed velocity

Then replacing with our values we have,

$9.75 = \frac{Bo^2 (3*10^8)}{(4\pi*10^{-7})}$

Re-arrange to find the magnetic Field B_0

$B_o = 2.86*10^{-7} T$

Therefore the amplitude of the magnetic field of this light is $B_o = 2.86*10^{-7} T$

6 0

3 years ago