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4 years ago

How many photons with 10 ev are required to produce 20 joules of energy?

1 answer:

6 0

One electron Volt (eV) is equal to 1.6 x 10^-19 Joules. Therefore, 10 eV is equal to 1.6 x 10^-18 Joules. In order to produce 20 Joules of energy from 10 eV photons, we would require 20 x 1/(1.6 x 10^-18) = 1.25 x 10^19 particles. This demonstrates that in the world of particle physics, the Joule is a massive energy unit relative to the commonly used electron Volt.

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Two asteroids collide and stick together. The first asteroid has mass of 1.50 × 104 kg andis initially moving at 0.77 × 103 m/s.

KengaRu [80]

Answer:

Magnitude 900m/s, direction 12.8° respect to the velocity of the first asteroid.

Explanation:

This is a perfectly inelastic collision, because the two asteroids stick together at the end. That means that the kinetic energy doesn't conserves, but the linear momentum does. But, since the velocities of the asteroids have different directions, we have to break down them in components. For convenience, we will take the direction of the first asteroid as x-axis, and its perpendicular direction (in the plane of the two velocity vectors) as y-axis. So, we have that:

$p_{1ox}+p_{2ox}=p_{fx}\\\\p_{2oy}=p_{fy}$

And, since $p=mv$ , we get:

$m_1v_{1o}+m_2v_{2o}\cos\theta=(m_1+m_2)v_{fx}\\\\m_2v_{2o}\sin\theta=(m_1+m_2)v_{fy}$

Solving for v_fx and v_fy, and calculating their values, we get:

$v_{fx}=\frac{m_1v_{1o}+m_2v_{2o}\cos\theta}{m_1+m_2}\\\\\implies v_{fx}=\frac{(1.50*10^{4}kg)(0.77*10^{3}m/s)+(2.00*10^{4}kg)(1.02*10^{3}m/s)\cos20\°}{1.50*10^{4}kg+2.00*10^{4}kg}=878m/s\\\\\\v_{fy}=\frac{m_2v_{2o}\sin\theta}{m_1+m_2}\\\\\implies v_{fy}=\frac{(2.00*10^{4}kg)(1.02*10^{3}m/s)\sin20\°}{1.50*10^{4}kg+2.00*10^{4}kg}=199m/s$

Now, the final speed can be calculated using the Pythagorean Theorem:

$v_f=\sqrt{v_{fx}^{2}+v_{fy}^{2}} \\\\\implies v_f=\sqrt{(878m/s)^{2}+(199m/s)^{2}}=900m/s$

And the direction $\beta=\arctan \frac{v_{fy}}{v_{fx}}\\ \\\implies \beta=\arctan\frac{199m/s}{878m/s}=12.8\°$ can be obtained using trigonometry:

$\beta=\arctan \frac{v_{fy}}{v_{fx}}\\ \\\implies \beta=\arctan\frac{199m/s}{878m/s}=12.8\°$

That means that the final velocity of the two asteroids has a magnitude of 900m/s and a direction of 12.8° with respect to the velocity of the first asteroid.

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3 years ago