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Nastasia [14]
3 years ago
8

How many neutrons and electrons are in ruthenium??? PLEASE HELP IM DROWNING IN WORK!!!!!

Physics
2 answers:
Andrei [34K]3 years ago
6 0
The Ruthenium Ru is a chemical element belonging to the Platinum group. If you look on the periodic table, his atomic number is Z=44. Atomic number represents too the number of Protons, and since an atom is electrically neutral, the number of protons is equal to the number of electrons. So Ruthenium got 44 electrons.
The mass number of the ruthenium based on the periodic table is A=101.
A = Z+N with N the number of neutrons. So N = A-Z = 101 - 44 = 57.
So the Ruthenium contains 57 neutrons.

Just for more information, the Ruthenium is commonly used in Jewelry and in electricity in case of severe wear resistance.

Hope this Helps! :)
Ierofanga [76]3 years ago
5 0
57 nuetrons and 44 elctrons just incase you need it there are 44 protons too.Hope this halps
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For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

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4) The potential energy at <u>point C</u> is zero

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6) The velocity of the roller coaster at <u>point C</u> is 19.91 m/s

1) The total energy for the roller coaster at the <u>initial point</u> can be found as follows:

E_{t} = KE_{i} + PE_{i}

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

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The<em> total energy</em> is:

E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J

Hence, the total energy for the roller coaster at the <u>initial point</u> is 19820 J.

   

2) The <em>potential energy</em> at point A is:

PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J

Then, the potential energy at <u>point A</u> is 19620 J.

3) The <em>kinetic energy</em> at point B is the following:

KE_{A} + PE_{A} = KE_{B} + PE_{B}

KE_{B} = KE_{A} + PE_{A} - PE_{B}

Since

KE_{A} + PE_{A} = KE_{i} + PE_{i}

we have:

KE_{B} = KE_{i} + PE_{i} - PE_{B} =  19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J

Hence, the kinetic energy at <u>point B</u> is 10010 J.

4) The <em>potential energy</em> at <u>point C</u> is zero because h = 0 meters.

PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J

5) The <em>kinetic energy</em> of the roller coaster at point C is:

KE_{i} + PE_{i} = KE_{C} + PE_{C}            

KE_{C} = KE_{i} + PE_{i} = 19820 J      

Therefore, the kinetic energy at <u>point C</u> is 19820 J.

6) The <em>velocity</em> of the roller coaster at point C is given by:

KE_{C} = \frac{1}{2}mv_{C}^{2}

v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s

Hence, the velocity of the roller coaster at <u>point C</u> is 19.91 m/s.

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