Answer:
Nuclear fusion
Explanation:
In fusion, many nuclei (the centers of atoms) combine together to make a larger one (which is a different element). ... A star's mass determines what other type of nucleosynthesis occurs in its core (or during explosive changes in its life cycle).
Answer: D
Explanation:
Atomic weight is measured by adding the number of protons and neutrons in an atom's nucleus. Argon's atomic number is 18 while potassium's is 19. This means that Argon will always have 18 protons while potassium will always have 19 protons.
To make the numbers easier to work with, round each atomic weight. We'll say the atomic weight of potassium is 39 and the atomic weight of argon is 40. To see how many neutrons each one has, I can set up a simple equation for each using the following equation:
Atomic weight = protons + neutrons
Potassium:
39 = 19 + N --> N = 20
Argon:
40 = 18 + N --> N = 22
An atom is defined by the number of protons it has, but the number of neutrons can vary. We call these isotopes, or atoms with the same number of protons but a different number of neutrons. As the math shows, argon typically has more neutrons per atom than potassium does.
Answer:
The frequency of the phonograph record is 0.2 Hz
Explanation:
The frequency of an object moving in uniform circular motion is the number of completed cycles the object makes in a specified time period
The given parameters of the phonograph record are;
The radius of the record = 0.15 m
The number of times the phonograph record rotates, n = 18 times
The time it takes the phonograph record to rotate the 18 times, t = 90 seconds
The frequency of the phonograph record, f = (The number of times the phonograph record rotates) ÷ (The time it takes the phonograph record to rotate the 18 times)
∴ The frequency of the phonograph record, f = n/t = 18/(90 s) = 0.2 Hz
The frequency of the phonograph record = 0.2 Hz.
Answer: 9.9%
Explanation: efficiency = (work output /work input) × 100
Note that, 1 kilocalorie = 4184 joules, hence 22kcal = 22× 4184 = 92048 joules.
Work output = 9200 j and work input = 92048 j
Efficiency = (9200/92048) × 100 = 0.099 × 100 = 9.9%
