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7nadin3 [17]
3 years ago
15

Is AG on the periodic table

Physics
1 answer:
Taya2010 [7]3 years ago
8 0
Ag on the periodic table is silver. 
The atomic number for silver is 47, atomic symbol is Ag, and the atomic weight is 107.9.
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Examine the scenario. Object A has 5 protons and 5 electrons. Object B has 5 protons and 7 electrons. Which option most accurate
Travka [436]
<span>Last choice on the list:
Object A has a net charge of 0 because the positive and negative
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Object B has a net charge of –2 because there is an imbalance of
charged particles (2 more negative electrons than positive protons).</span>
4 0
3 years ago
Im bad at work problems can any one help with this problem ?
lyudmila [28]
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8    7,680    1,010
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12  14,640  1,790
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4 0
3 years ago
Which of the following surfaces reflects the most light
Gre4nikov [31]

Answer:

D

Explanation:

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3 0
2 years ago
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The​ half-life of a certain radioactive substance is 12 hours. There are 19 grams present initially. a. Express the amount of su
neonofarm [45]

Answer:

(a) N=19\times e^{-\lambda t}

(b) 15 hours

Explanation:

half life, T = 12 hours

No = 19 g

(a) Let N be the amount remaining after time t.

Let λ be the decay constant.

\lambda =\frac {0.6931}{T}

The equation of radioactivity used here is given by

N=N_{o}e^{-\lambda t}

N=19\times e^{-\lambda t}

(b) N = 8 gram

Substitute the values in above equation

\lambda =\frac {0.6931}{12}

λ = 0.0577 per hour

So, 8=19\times e^{-0.577t}

e^{-0.0577t}=0.421

Take natural log on both the sides

- 0.0577 t = - 0.865

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4 0
3 years ago
An infinitely long straight wire has a uniform linear charge density of Derive the 4. equation for the electric field a distance
marshall27 [118]

Answer:

E = \frac{\lambda}{2\pi \epsilon_0 r}

Explanation:

Let the linear charge density of the charged wire is given as

\frac{q}{L} = \lambda

here we can use Gauss law to find the electric field at a distance r from wire

so here we will assume a Gaussian surface of cylinder shape around the wire

so we have

\int E. dA = \frac{q}{\epsilon_0}

here we have

E \int dA = \frac{\lambda L}{\epsilon_0}

E. 2\pi r L = \frac{\lambda L}{\epsilon_0}

so we have

E = \frac{\lambda}{2\pi \epsilon_0 r}

4 0
2 years ago
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