Answer:
Stable atom
Explanation:
A stable atom is one that has a balanced nuclear inter-particle force reaction as such the binding energy of a stable atom is sufficient to permanently keep the nucleus as one unit. Examples of a stable atom are the atoms of monoisotopic elements such as fluorine, sodium, iodine, gold, aluminium, and cobalt.
In a stable atom the expected number of proton, neutron, and electron are present while in an unstable atom or radioactive atom, there are more than the expected number of neutrons or protons, such that the internal energy of the nucleus is excessive and more than the binding energy, which can lead to radioactive decay.
The answer is 17 m because you have to add the 15 m and the 8 m together to get the answer so it will be like this 17x17 = 15x15 + 8x8 got it?
Answer:
Explanation:
The relation between time period of moon in the orbit around a planet can be given by the following relation .
T² = 4 π² R³ / GM
G is gravitational constant , M is mass of the planet , R is radius of the orbit and T is time period of the moon .
Substituting the values in the equation
(.3189 x 24 x 60 x 60 s)² = 4 x 3.14² x ( 9380 x 10³)³ / (6.67 x 10⁻¹¹ x M)
759.167 x 10⁶ = 8.25 x 10²⁰ x 39.43 / (6.67 x 10⁻¹¹ x M )
M = .06424 x 10²⁵
= 6.4 x 10²³ kg .
Answer:
67
Explanation:
- The atomic number (Z) of an atom is equal to the number of protons in the nucleus
- The mass number (A) of an atom is equal to the sum of protons and neutrons in the nucleus
Therefore, calling p the number of protons and n the number of neutrons, for element X we have:
Z = p = 23
A = p + n = 90
Substituting p=23 into the second equation, we find the number of neutrons:
n = 90 - p = 90 - 23 = 67
Answer:
<h2>A.
6pF</h2>
Explanation:
If unknown capacitance C1, C2, C3 and C4 are connected in series to one another, their equivalent capacitance of the circuit will be expressed as shown
![\frac{1}{C_t} = \frac{1}{C_1} +\frac{1}{C_2} +\frac{1}{C_3} +\frac{1}{C_4} \\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BC_t%7D%20%3D%20%5Cfrac%7B1%7D%7BC_1%7D%20%2B%5Cfrac%7B1%7D%7BC_2%7D%20%2B%5Cfrac%7B1%7D%7BC_3%7D%20%2B%5Cfrac%7B1%7D%7BC_4%7D%20%5C%5C)
Given the capacitance's 3.0 pF, 2.0 pF, 5.0 pF and X pF connected in series to each other. If the equivalent capacitance of the circuit is 0.83 pF, then to get X, we will apply the formula above;
![\frac{1}{0.83} = \frac{1}{3.0} +\frac{1}{2.0} +\frac{1}{5.0} +\frac{1}{C_4} \\\\\\1.205 = 0.333+0.5+0.2+\frac{1}{C_4} \\\\1.205 = 1.033 + \frac{1}{C_4} \\\\\frac{1}{C_4} = 1.205-1.033\\\\\frac{1}{C_4} = 0.172\\\\C_4 = \frac{1}{0.172}\\ \\C_4 = 5.8pF\\\\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B0.83%7D%20%3D%20%5Cfrac%7B1%7D%7B3.0%7D%20%2B%5Cfrac%7B1%7D%7B2.0%7D%20%2B%5Cfrac%7B1%7D%7B5.0%7D%20%2B%5Cfrac%7B1%7D%7BC_4%7D%20%5C%5C%5C%5C%5C%5C1.205%20%3D%200.333%2B0.5%2B0.2%2B%5Cfrac%7B1%7D%7BC_4%7D%20%5C%5C%5C%5C1.205%20%3D%201.033%20%2B%20%5Cfrac%7B1%7D%7BC_4%7D%20%5C%5C%5C%5C%5Cfrac%7B1%7D%7BC_4%7D%20%20%3D%201.205-1.033%5C%5C%5C%5C%5Cfrac%7B1%7D%7BC_4%7D%20%20%3D%200.172%5C%5C%5C%5CC_4%20%3D%20%5Cfrac%7B1%7D%7B0.172%7D%5C%5C%20%5C%5CC_4%20%3D%205.8pF%5C%5C%5C%5C)
C₄ ≈ 6pF
Hence the value of the X capacitor is approximately 6pF