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2 years ago

Donde esta la orilla del universo?

Physics

1 answer:

irinina [24]2 years ago

5 0

Answer:

<em>El universo no tiene un borde finito</em>, continúa y se expande sin fin. La distancia adecuada desde la Tierra hasta el borde del universo observable es de aproximadamente 46.5 mil millones de años luz o 4.40 × 10 ^ 26 metros en cualquier dirección.

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3 years ago

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Explanation:

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The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re

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Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

$\Delta x = \Delta y$

Where:

$\Delta x$ - Range of the bullet, measured in meters.

$\Delta y$ - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

$K_{1} = U_{2}$

Where:

$K_{1}$ - Kinetic energy at point 1, measured in joules.

$U_{1}$ - Gravitational potential energy at point 2, measured in joules, and:

$U_{2} = m\cdot g \cdot \Delta y$

Where:

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

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

The maximum height is now cleared:

$K_{1} = m\cdot g \cdot \Delta y$

$\Delta y = \frac{K_{1}}{m\cdot g}$

If $K_{1} = 1066\,J$ , $m = 0.25\,kg$ and $g = 9.81\,\frac{m}{s^{2}}$ , the maximum height is now computed:

$\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

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3 years ago

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Answer:

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Explanation:

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3 years ago

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Part 1)

Answer:

Explanation:

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$V = E(1 - e^{-t/RC})$

so we will have

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here we know that

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$t = 1.32 s$

Part b)

Answer:

The time will increase.

Explanation:

As we know that on increasing the value of the resistance the the product of the resistance and capacitance will increase so the time will increase to get the above voltage.

Part c)

Answer:

The capacitor discharges through a very low resistance (the lamp filled with ionized gas), and so the discharge time constant is very short. Thus the flash is very brief.

Explanation:

Since the lamp resistance is very small so the energy across the lamp will totally lost in very short interval of time

Part d)

Answer:

Once the lamp has flashed, the stored energy in the capacitor is gone, and there is no source of charge to maintain the lamp current. The lamp "goes out", the lamp resistance increases, and the capacitor starts to recharge. It charges again and the process will repeat.

Explanation:

Since we know that the battery is connected to the given system so after whole energy of capacitor is flashed out it is again charged by the battery and the process will continue

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3 years ago