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

- Tin:

Atomic number = 50

Mass number = 119

Number of neutrons = mass number - atomic number = 119 - 50

Number of neutrons = 69

- Antimony(Sb):

Atomic number = 51

Mass number = 122

Number of neutrons = mass number - atomic number = 122 - 51

Number of neutrons = 71

- Tellurium(Te):

Atomic number = 52

Mass number = 128

Number of neutrons = mass number - atomic number = 128 - 52

Number of neutrons = 76

- Iodine(I):

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).

7 0

3 years ago

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A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The ang

olga55 [171]

Answer:

9.22 s

Explanation:

One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad

Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:

For the child: $s_c = \omega_ct = 0.233t$

For the horse: $s_h = s_0 + a_ht^2/2 = 1.57 + 0.0136t^2/2 = 1.57 + 0.0068t^2$

For the child to catch up with the horse, they must cover the same angular distance within the same time t:

$s_c = s_h$

$0.233t = 1.57 + 0.0068t^2$

$0.0068t^2 - 0.233t + 1.57 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{0.233\pm \sqrt{(-0.233)^2 - 4*(0.0068)*(1.57)}}{2*(0.0068)}$

$t= \frac{0.233\pm0.11}{0.0136}$

t = 25.05 or t = 9.22

Since we are looking for the shortest time we will pick t = 9.22 s

6 0

3 years ago

suppose a vacuum cleaner use 120 j of electrical energy. If 45 j are used to pull air into the vacuum cleaner, how efficient is

Mice21 [21]

It depends on how much time 45 j will go, so if you told me that it went through 45 j per minute it will go through for 2 minutes so not efficient unless it went through 45 j per hour then it is efficient.

8 0

3 years ago