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3 years ago

Why is a human powered generator better than the sun (solar cells)?

Physics

1 answer:

zheka24 [161]3 years ago

6 0

Answer:

we can not use the suns energy too effectively in power cells and with human power we can generate more energy

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nika2105 [10]

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3 years ago

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Maslowich

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3 years ago

Read 2 more answers

A 54.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 126 m/s from the top of a cliff

Nikolay [14]

Answer:

a) The initial total mechanical energy of the projectile is 498556.296 joules.

b) The work done on the projectile by air friction is 125960.4 joules.

c) The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

Explanation:

a) The system Earth-projectile is represented by the Principle of Energy Conservation, the initial total mechanical energy ( $E$ ) of the project is equal to the sum of gravitational potential energy ( $U_{g}$ ) and translational kinetic energy ( $K$ ), all measured in joules:

$E = U_{g} + K$ (Eq. 1)

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$E = m\cdot g\cdot y + \frac{1}{2}\cdot m\cdot v^{2}$ (Eq. 1b)

Where:

$m$ - Mass of the projectile, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$y$ - Initial height of the projectile above ground, measured in meters.

$v$ - Initial speed of the projectile, measured in meters per second.

If we know that $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y = 132\,m$ and $v = 126\,\frac{m}{s}$ , the initial mechanical energy of the earth-projectile system is:

$E = (54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (132\,m)+\frac{1}{2}\cdot (54\,kg)\cdot \left(126\,\frac{m}{s} \right)^{2}$

$E = 498556.296\,J$

The initial total mechanical energy of the projectile is 498556.296 joules.

b) According to this statement, air friction diminishes the total mechanical energy of the projectile by the Work-Energy Theorem. That is:

$W_{loss} = E_{o}-E_{1}$ (Eq. 2)

Where:

$E_{o}$ - Initial total mechanical energy, measured in joules.

$E_{1}$ - FInal total mechanical energy, measured in joules.

$W_{loss}$ - Work losses due to air friction, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$W_{loss} = E_{o}-K_{1}-U_{g,1}$

$W_{loss} = E_{o} -\frac{1}{2}\cdot m\cdot v_{1}^{2}-m\cdot g\cdot y_{1}$ (Eq. 2b)

Where:

$m$ - Mass of the projectile, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$y_{1}$ - Maximum height of the projectile above ground, measured in meters.

$v_{1}$ - Current speed of the projectile, measured in meters per second.

If we know that $E_{o} = 498556.296\,J$ , $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{1} = 297\,m$ and $v_{1} = 89.3\,\frac{m}{s}$ , the work losses due to air friction are:

$W_{loss} = 498556.296\,J -\frac{1}{2}\cdot (54\,kg)\cdot \left(89.3\,\frac{m}{s} \right)^{2} -(54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (297\,m)$

$W_{loss} = 125960.4\,J$

The work done on the projectile by air friction is 125960.4 joules.

c) From the Principle of Energy Conservation and Work-Energy Theorem, we construct the following model to calculate speed of the projectile before it hits the ground:

$E_{1} = U_{g,2}+K_{2}+1.5\cdot W_{loss}$ (Eq. 3)

$K_{2} = E_{1}-U_{g,2}-1.5\cdot W_{loss}$

Where:

$E_{1}$ - Total mechanical energy of the projectile at maximum height, measured in joules.

$U_{g,2}$ - Potential gravitational energy of the projectile, measured in joules.

$K_{2}$ - Kinetic energy of the projectile, measured in joules.

$W_{loss}$ - Work losses due to air friction during the upward movement, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$\frac{1}{2}\cdot m \cdot v_{2}^{2} = E_{1}-m\cdot g\cdot y_{2}-1.5\cdot W_{loss}$ (Eq. 3b)

$m\cdot v_{2}^{2} = 2\cdot E_{1}-2\cdot m \cdot g \cdot y_{2}-3\cdot W_{loss}$

$v_{2}^{2} = 2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m}$

$v_{2} = \sqrt{2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m} }$

If we know that $E_{1} = 372595.896\,J$ , $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{2} =0\,m$ and $W_{loss} = 125960.4\,J$ , the final speed of the projectile is:

$v_{2} =\sqrt{2\cdot \left(\frac{372595.896\,J}{54\,kg}\right)-2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0\,m)-3\cdot \left(\frac{125960.4\,J}{54\,kg}\right) }$