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3 years ago

A)what are oxidation numbers? b) what useful functions do oxidation numbers serve

1 answer:

5 0

A) indicate the general distribution of electrons among bonded atoms in molecular compounds
B)useful in naming compounds, writing formulas, and balancing chemical equations
may be used to determine the simplest chemical formula

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Which of the following is necessary for the greenhouse effect?

Mice21 [21]

It is the atmosphere

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3 years ago

What metric units are used to report the mass of an object

sladkih [1.3K]

The unit used to measure weight in the metric system is the gram.

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4 years ago

Three identical particles, q1, q2, and q3, each with charge q = 5.00 μC, are placed along a circle of radius r = 2.00 m at angle

wariber [46]

Answer:

11250 N/C

Direction: 0 deg counterclockwise from positive x-axis

Explanation:

$q$ = magnitude of charge on each particle = 5 μC = 5 x 10⁻⁶ C

$r$ = distance of each particle from center of circle = 2 m

$E$ = Magnitude of electric field at the center by each particle

Magnitude of electric field at the center by each particle is given as

$E = \frac{kq}{r^{2} }$

inserting the values

$E = \frac{(9\times10^{9} )(5\times10^{-6})}{2^{2} }\\E = 11.25\times10^{3} NC^{-1}$

From the diagram , we see that being equal and opposite, the electric fields due to charge q₁ and q₃ cancel out.

So net electric field at center is only due to charge q₂ direction towards positive x-direction

$E_{res}$ = Resultant electric field = 11250 N/C

Direction: 0 deg counterclockwise from positive x-axis

7 0

3 years ago

What is the spring coefficient of a spring that stores 4000 joules of energy

Verizon [17]

Answer:

222.2N/m

Explanation:

Given parameters:

Elastic potential energy = 4000J

Extension = 6m

Unknown:

Spring coefficient = ?

Solution:

The elastic potential energy is the energy stored within a string.

It is expressed as;

EPE = $\frac{1}{2}$ k e²

k is the spring constant

e is the extension

4000 = $\frac{1}{2}$ x k x 6²

8000 = 36k

k = 222.2N/m

6 0

3 years ago

A +8.75 μC point charge is glued down on a horizontal frictionless table. It is tied to a -6.50 μC point charge by a light, nonc

lisov135 [29]

(a) The tension on the wire when the two charges have opposite signs is 383.5 N.

(b) The tension on the wire if both charges were negative is 3.640.25 N.

The given parameters;

first charge, q₁ = 8.75 μC
second charge, q₂ = -6.5 μC
electric field, E = 1.85 x 10⁸ N/C
distance between the two charges, r = 2.5 cm

(a)

The attractive force between the charges is calculated as follows;

$F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = -819 \ N$

The force on the negative charge due to the electric field is calculated as follows;

$F_2 = Eq_2\\\\F_2 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_2 = 1202.5 \ N$

The tension on the wire is the resultant of the two forces and it is calculated as follows;

$T = F_2 + F_1\\\\T = 1202.5 - 819\\\\T = 383.5 \ N$

(b) when the two charges are negative

The repulsive force between the two charges is calculated as follows;

$F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (-8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = 819 \ N$

The force on the first negative charge due to the electric field is calculated as follows;

$F_2 = Eq_1\\\\F_2 = (1.85 \times 10^8)\times (8.75 \times 10^{-6})\\\\F_2 = 1618.75 \ N$

The force on the second negative charge due to the electric field is calculated as follows;

$F_3 = Eq_2\\\\F_3 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_3 = 1202.5 \ N$

The tension on the wire is the resultant of the three forces and it is calculated as follows;

$T= F_1 + F_2 + F_3\\\\T= 819 + 1618.75 + 1202.5\\\\T = 3,640.25 \ N$

Learn more here:brainly.com/question/19565286

5 0

3 years ago