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Answer:
15 m/s or 1500 cm/s
Explanation:
Given that
Speed of the shoulder, v(h) = 75 cm/s = 0.75 m/s
Distance moved during the hook, d(h) = 5 cm = 0.05 m
Distance moved by the fist, d(f) = 100 cm = 1 m
Average speed of the fist during the hook, v(f) = ? cm/s = m/s
This can be solved by a very simple relation.
d(f) / d(h) = v(f) / v(h)
v(f) = [d(f) * v(h)] / d(h)
v(f) = (1 * 0.75) / 0.05
v(f) = 0.75 / 0.05
v(f) = 15 m/s
Therefore, the average speed of the fist during the hook is 15 m/s or 1500 cm/s
Answer:
B
Explanation:
the 3 electrons makes it neutral
The correct answer is:

Let's see why.
1 amu corresponds to the mass of the proton, which is:

if we convert this into energy, using Einstein equivalence between mass and energy, we find:

Now we can convert it into electronvolts:

So, 1 amu = 934 MeV. Therefore, 3 amu corresponds to 3 times this value: