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zlopas [31]
3 years ago
9

The sun, like all stars, releases energy through nuclear fusion. In this problem, you will find the total number of fusion react

ion events that occur inside the sun every second. You will be considering the proton-proton chain, in which four hydrogen nuclei are converted into a helium nucleus and two positrons. The net reaction for the proton-proton chain is
411H? 42He+2?+ .

To find the energy released by this reaction, you will need the following mass data:

mass of 11H=1.007825u

and

mass of 42He=4.002603u .

Using the masses of the neutral atoms in your calculation accounts for the energy released by the annihilation of the positrons with electrons, so you can work this problem without reference to the positrons or their rest mass.

a. What is the total energy Q released in a single fusion reaction event for the equation given in the problem introduction?
Physics
1 answer:
sattari [20]3 years ago
4 0

Answer:

Explanation:

DetaM=4 x 1.02875 - 4.002603

DetaM= 0.028697u

Using E= mc²

= 0.028697 x 1.49x*10^-10

= 4.2x10^-12J

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If the Sun were to stop shining, Earth would stop receiving heat and light within about eight minutes. Earth, however, would rem
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If<span> The </span>Sun<span> Went Out, How Long </span>Could<span> Life On </span>Earth<span> Survive? ... (which is actually physically impossible), the </span>Earth would stay<span> warm—at least ... from the planet's core </span>would<span> equal the</span>heat<span> that the </span>Earth<span> radiates into space, ... Photosynthesis </span>would<span> halt immediately, and </span>most<span> plants</span>would<span> die </span>in<span> a few </span>weeks<span>.</span>
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A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15
Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

T-mg = m\frac{v^2}{R}

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

m \frac{v^2}{R} is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

T=m\frac{v^2}{R}

And by substituting the numerical values, we find

T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

T+mg=m\frac{v^2}{R}

And substituting the numerical values and re-arranging it, we find

T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

T=0\\mg = m\frac{v^2}{R}

and re-arranging the equation, we find

v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s

7 0
3 years ago
A falling skydiver has a mass of 105 kg. What is the magnitude of the skydiver's acceleration when the upward force of air resis
LenKa [72]
F-free = m*g - F_air = m*a

F_air = 1.2 * m

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a = 9.3 m/s

Hope this helps
7 0
3 years ago
1. The horizontal and vertical components of a projectile's velocity are
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The horizontal and vertical components of a projectile's velocity are independent of each other.

Answer: Option C

<u>Explanation:</u>

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Even it can seen that when the horizontal components of velocity is constant, then there will be change in the vertical components of velocity leading to free fall projectile path.

And in the absence of gravity, there will be change in the horizontal components of velocity with zero vertical component of velocity. Thus, the horizontal and the vertical components of a projectile’s velocity are seemed to be independent of each other.

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