Answer:
Atoms of tellurium (Te) have the greatest average number of neutrons equal to 76.
Explanation:
In the periodic table, Elements are represented with their respected symbols. Above the symbol is the elements atomic number which is equal to the number of protons in each atom. Below the symbol is the mass number of that element which is roughly equal to the sum of neutrons and protons of that atom.
To calculate the number of neutrons we can take the difference of Atomic number and mass number:
Number of neutrons = mass number - atomic number
<u>- Tin:</u>
Atomic number = 50
Mass number = 119
Number of neutrons = mass number - atomic number = 119 - 50
Number of neutrons = 69
<u>- Antimony(Sb):</u>
Atomic number = 51
Mass number = 122
Number of neutrons = mass number - atomic number = 122 - 51
Number of neutrons = 71
<u>- Tellurium(Te):</u>
Atomic number = 52
Mass number = 128
Number of neutrons = mass number - atomic number = 128 - 52
Number of neutrons = <u>76</u>
<u>- Iodine(I):</u>
Atomic number = 53
Mass number = 127
Number of neutrons = mass number - atomic number = 127 - 53
Number of neutrons = 74
Here, the greatest number of neutrons is for the atoms of Tellurium(Te).
Answer:
The charge on the third object is − 21.7nC
Explanation:
From Gauss's Law
Φ = Q/ε₀
where;
Φ is the total electric flux through the shell = − 533 N⋅m²/C
Q is the total charge Q in the shell = ?
ε₀ is the permittivity of free space = 8.85 x 10⁻¹²
From this equation; Φ = Q/ε₀
Q = Φ * ε₀ = − 533 * 8.85 x 10⁻¹²
Q = −4.7 X 10⁻⁹ C = -4.7nC
Q = q₁ + q₂ + q₃
− 4.7nC = − 14.0 nC + 31.0 nC + q₃
− 4.7nC − 17nC = q₃
− 21.7nC = q₃
Therefore, the charge on the third object is − 21.7nC
The new oscillation frequency of the pendulum clock is 1.14 rad/s.
The given parameters;
- <em>Mass of the pendulum, = M </em>
- <em>Length of the pendulum, = L</em>
- <em>Initial angular speed, </em>
<em> = 1 rad/s</em>
The moment of inertia of the rod about the end is given as;

The moment of inertia of the rod between the middle and the end is calculated as;
![I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} = \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}](https://tex.z-dn.net/?f=I_f%20%3D%20%5Cint%5Climits%5EL_%7BL%2F2%7D%20%7Br%5E2%5Cfrac%7BM%7D%7BL%7D%20%7D%20%5C%2C%20dr%20%3D%20%5Cfrac%7BM%7D%7B3L%7D%20%5Br%5E3%5D%5EL_%7BL%2F2%7D%20%3D%20%20%5Cfrac%7BM%7D%7B3L%7D%20%5BL%5E3%20-%20%5Cfrac%7BL%5E3%7D%7B8%7D%20%5D%20%3D%20%5Cfrac%7BM%7D%7B3L%7D%20%5B%5Cfrac%7B7L%5E3%7D%7B8%7D%20%5D%3D%20%5Cfrac%7B7ML%5E2%7D%7B24%7D)
Apply the principle of conservation of angular momentum as shown below;

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.
Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129
Transverse, I think. I may be wrong.