Answer:
option a.
Explanation:
We can think of an atom as a nucleus (where the protons and neutrons are) and some electrons orbiting it.
We also know that the mass of an electron is a lot smaller than the mass of a proton or the mass of an electron.
So, if all the protons and electrons of an atom are in the nucleus, we know that most of the mass of an atom is in the nucleus of that atom.
Then we define the mass number, which is the total number of protons and neutrons in an atom. Such that the mass of a proton (or a neutron) is almost equal to 1u
Then if we define A as the total number of protons and neutrons, and each one of these weights about 1u
(where u = atomic mass unit)
Then the weight of the nucleus is about A times 1u, or:
A*1u = A atomic mass units.
Then the correct option is:
The mass of the nucleus is approximately EQUAL to the mass number multiplied by __1__ Atomic Mass unit.
option a.
<span>ripple factor can be reduced by increasing the value of the load resistor (which means reducing the load of the circuit)</span>
We can substitute the given values into the equation for T, given the surrounding temperature T0 = 0, initial temperature T1 = 140, constant k = -0.0815, and time t = 15 minutes.
T = 0 + (140 - 0)e^(-0.0815*15) = 140e^(-1.2225) = 41.23°F
Answer:
B. the stars to come back to the same positions in the sky.
Explanation:
In fact, the solar day is equivalent to more than a rotation, because when the point has turned completely, it is not, as it should, in the same position with respect to the Sun.
The reason for this is that while performing the rotation, the Earth simultaneously moved following its orbit around the Sun.
When the reference point completed its rotation, the Earth already moved almost 2,500,000 km., So that to see the Sun again it will be necessary to turn a little more.
Solar day is more than a rotation. The sidereal or sidereal day, commonly used by astronomers, is also based on the rotation of the Earth; but in this case a distant star is taken as a reference (sidereal comes from the Latin sidus which means "star").
Answer:
d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero
Explanation:
Affirmations
a) true. The orbits are accurate in the Bohr model and probabilistic in quantum mechanics
b) True. If both give the same results and use the same quantum number (n)
c) True. If in angular momentum it is quantized, in the Bohr model too but it does not justify it
d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero
