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slamgirl [31]
3 years ago
12

assume the suns total energy output is 4.0 * 10^26 watts, and 1 watt is 1 joule/second. assume 4.3 * 10^-12 J is released from e

ach p-p chain of reactions. use the energy yield from the proton-proton chain to determine how many proton-proton chain reactions must be happening each second in the solar core
Physics
1 answer:
Dmitry [639]3 years ago
3 0

Answer:

9.3\cdot 10^{37}

Explanation:

Power is defined as the energy produced (E) per unit of time (t):

P= \frac{E}{t}

This means that the energy produced in the Sun each second (1 s), given the power P=4.0\cdot 10^{26}W, is

E=Pt=(4.0\cdot 10^{26}W)(1s )=4.0\cdot 10^{26} J

Each p-p chain reaction produces an amount of energy of

E_1 = 4.3\cdot 10^{-12} J

in order to get the total number of p-p chain reactions per second, we need to divide the total energy produced per second by the energy produced by each reaction:

n=\frac{E}{E_1}=\frac{4.0\cdot 10^{26} J}{4.3\cdot 10^{-12} J}=9.3\cdot 10^{37}

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Answer:

the time taken t is 9.25 minutes

Explanation:

Given the data in the question;

The initial charge on the supercapacitor = 2.1 × 10³ mV = 2.1 V

now, every minute, the charge lost is 9.9 %  

so we need to find the time for which the charge drops below 800 mV or 0.8 V

to get the time, we can use the formula for compound interest in basic mathematics;

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A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet
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Answer:

v  = 2.8898 \frac{m}{s}

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

E_{ep} = \frac{1}{2} k (\Delta x)^2

where \Delta x is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

\vec{F} = - k \Delta \vec{x}

as we need 9.12 N to compress 4.60 cm, this means:

k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}

k = 198.26 \ \frac{ N}{m}

So, the elastic energy of the compressed spring is:

E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2

E_{ep} = 0.209759 \ Joules

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

\Delta E_{gp} = m g \Delta h

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m

\Delta E_{gp} = 0.00224 \ Joules

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

E_k = \frac{1}{2} m v^2

so, for the pellet will be

E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2

<h3>All together</h3>

By conservation of energy, we know:

E_{ep} = \Delta E_{gp} + E_k

0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2

So

\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2  = 0.209759 \ Joules - 0.00224 \ Joules

\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2  = 0.207519 \ Joules

v  = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }

v  = 2.8898 \frac{m}{s}

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