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olganol [36]
3 years ago
5

What can happen to an electron when sunlight hits it? select all that apply. select all that apply. it can drop down to a lower

electron shell. it can move out to a higher electron shell. it can collide with the nucleus. it can stay in its original shell?
Physics
2 answers:
Bingel [31]3 years ago
8 0
An electron gains energy when powered by solar energy, so the first option of it dropping to a lower electron shell isn't possible. It can however, move out to a higher electron shell. It cannot collide with the nucleus, nor would it remain in its original shell, as it has gained energy. 
vlabodo [156]3 years ago
4 0
There are two possible answers:
<span>- it can move out to a higher electron shell
- </span><span> it can stay in its original shell
</span><span>
In fact, sunlight consists of photons. When sunlight hits an electron, the electron can absorbs a photon, so it gains energy: as a result, the electron can move to a higher electron shell, which corresponds to a high energy level in the atom, if the energy given by the photon is at least equal to the energy difference between the two levels. However, if the photon energy is not large enough, the electron will stay in the same shell.</span>
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I need Help ASAP, and actually help, not just for the points.
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750 force

Explanation:

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8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s
allsm [11]

Answer:

a) \omega = 50\,\frac{rad}{s}, b) \omega = 0\,\frac{rad}{s}

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

\tau = F \cdot r

\tau = m\cdot a \cdot r

\tau = m \cdot \alpha \cdot r^{2}

Where \alpha is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}

\alpha = -3\,\frac{rad}{s^{2}}

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)

\omega = 50\,\frac{rad}{s}

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}

t = 66.667\,s

Since t > 66.667\,s, then the angular velocity is equal to zero. Therefore:

\omega = 0\,\frac{rad}{s}

7 0
4 years ago
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