Well, it is also known as the transmission of fluid pressure. So, it is a principle in fluid mechanics that says that pressure exerted anywhere in a confined incompressible fluid is transmitted equally in all directions throughout the fluid such as the the pressure variations.
<span> In radioactive decay, an unstable atomic nucleus emits particles or radiation and converts to a different atomic nucleus. If the new nucleus is unstable, it will decay again, until eventually, a stable nucleus is formed. Such a sequence of nuclear decays forms a decay series.
The half-life of a radioactive substance is the time required for half of the atoms of a radioactive isotope to decay. If you have, say, 1 million atoms of a specific isotope in a sample, the time required for 500,000 of those atoms to decay is the half-life of that specific isotope. If you have 50 atoms of that isotope, 25 atoms will decay in the same amount of time.
Because the half-life is fixed for a specific isotope, it can be used to date objects. You compare the decay rate of an old object with the decay rate of a fresh sample. Nuclear decay is a first-order process and can be described by a specific mathematical equation, which depends on the decay rate and the half-life. Knowing those values, you can work back and determine the age of an object, as compared with a standard sample. Old objects will not have as much of a radioactive isotope in them as new objects, since the isotopes will have decayed over time in the old object.</span>
It is a reactant, as it, with water and CO2 reacts to form the products.
Answer:
1.) 274.5v
2.) 206.8v
Explanation:
1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.
The potential difference and charge across EACH capacitor will be
V = Voe
Where Vo = initial voltage
e = natural logarithm = 2.718
For the first capacitor 2.50 µF,
V = Vo × 2.718
746 = Vo × 2.718
Vo = 746/2.718
Vo = 274.5v
To calculate the charge, use the below formula.
Q = CV
Q = 2.5 × 10^-6 × 274.5
Q = 6.86 × 10^-4 C
For the second capacitor 6.80 µF
V = Voe
562 = Vo × 2.718
Vo = 562/2.718
Vo = 206.77v
The charge on it will be
Q = CV
Q = 6.8 × 10^-6 × 206.77
Q = 1.41 × 10^-3 C
B.) Using the formula V = Voe again
165 = Vo × 2.718
Vo = 165 /2.718
Vo = 60.71v
Q = C × 60.71
Q = C
