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Rudiy27
3 years ago
10

An object becomes charged when the atoms in the object gain or lose

Physics
1 answer:
ICE Princess25 [194]3 years ago
5 0
The answer is c......
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Answer:

We know that in the decay process the sum of molecular number as well as molecular weight should be constant.

The following three reaction are as follows

1 .

Alpha decay  of parent nuclide

_{92}^{235}\textrm{U} \rightarrow _{90}^{231}\textrm{Th}+_{2}^{4}\textrm{alpha }

The molecular number of alpha particle is 2 and molecular weight is 4 g/mol.

2.

Beta decay of daughter nuclide

_{90}^{231}\textrm{Th}\rightarrow _{91}^{231}\textrm{Pa}+_{-1}^{0}\textrm{beta}+v

v is the neutrino emission,The charge on the beta particle is zero.

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Alpha decay

_{91}^{231}\textrm{Pa}\rightarrow_{89}^{227}\textrm{Pa}+_{2}^{4}\textrm{alpha}

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Of the superalloys, the cobalt-base types are the easiest to braze. True or False?
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It is false thats my answer


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4 years ago
a force 2.4E2 N exists between a positive charge of 8E-5 C and a positive charge of 3E-5 C. What distance separates the charges?
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The distance between the two charges is 0.3 m

Explanation:

The electrostatic force between two charged objects is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

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q_1, q_2 are the charges of the two objects

r is the separation between the two charges

In this problem, we are given the following:

q_1 = 8\cdot 10^{-5} C

q_2 = 3\cdot 10^{-5} C

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Therefore, we can rearrange the equation to solve for r, the distance between the two charges:

r=\sqrt{\frac{kq_1 q_2}{F}}=\sqrt{\frac{(8.99\cdot 10^9)(8\cdot 10^{-5})(3\cdot 10^{-5})}{2.4\cdot 10^2}}=0.3 m

Learn more about electrostatic force:

brainly.com/question/8960054

brainly.com/question/4273177

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