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3 years ago

5

An oxygen atom picks up two additional, free floating electrons. Is the charge of the newly formed oxygen ion positive, negative

or neutral?

2 answers:

Alex Ar [27]3 years ago

6 0

Answer:The newly formed oxygen ion is negatively charged.

Explanation:

Cation : These are species of ions which carries positive charge that is number of protons are greater than the number of electrons.

Anion:These are species of ions which carries negative charge that is number of electrons are greater than the number of protons.

$O+2e^-\rightarrow O^{2-}$

When an oxygen atom gains two electrons it develops the charge of negative charge and forms anion.

Vinvika [58]3 years ago

5 0

I believe it would be negative as electrons carry negative energy

hope this helps :)

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nalin [4]

1. make good decisions

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answered these on edge. also, you marked this as physics even though its english. might wanna watch out for that lol, so youll get a quicker answer

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3 years ago

Read 2 more answers

If temp. gets cold resistance in thermostat increases so voltage across it increases AND LED lights brighter. I understand every

Valentin [98]

Normally, when something gets colder, its electrical resistance gets smaller. This is true of component-A in the drawing ... a simple resistor.

The component labeled 'B' has a strange and unusual symbol, and it's not a simple resistor. It's a "thermistor". The word "thermal" always has something to do with heat, and "thermistor" comes from "thermal resistor. These things can be manufactured either way ... using different materials, a thermistor can be manufactured so that its resistance goes UP, or goes DOWN, or doesn'tchange when it gets colder. I'm pretty sure that's what's going on here.

When this circuit gets colder, resistance-A gets smaller, but resistance-B either gets bigger OR doesn't change. Either way, the voltage across B increases. Since the LED is connected directly across B, the current through it depends on that voltage, so the LED gets more current, and becomes brighter, when A and B both get colder.

This circuit could actually be a very useful device. If you took out the LED and put a voltmeter in its place, then the reading on the voltmeter would tell you the temperature of wherever you put the two components A and B.

5 0

3 years ago

A 10 g bullet moving horizontally with a speed of 2000 m/s strikes and passes through a 4.0 kg block moving with a speed of 4.2

Masteriza [31]

Answer:

The answer is "512 J".

Explanation:

bullet mass $m_1 = 10 g= 10^{-2} \ kg\\\\$

initial speed $u_1 = 2\ \frac{Km}{s}= 2000\ \frac{m}{s}\\\\$

block mass $m_2 = 4\ Kg$

initial speed $v_2 =-4.2 \frac{m}{s}$

final speed $v_2= 0$

Let $v_1$ will be the bullet speed after collision:

throughout the consevation the linear moemuntum

$\to M_1V_1+m_2v_2=M_1U_1+m_2u_2\\\\\to (10^{-2} kg) V_1 +0 = (10^{-2} kg)(2000 \frac{m}{s}) + (4 \ kg)(-4.2 \frac{m}{s}) \\\\\ \to 10^{-2} v_1 = 20 -16.8\\\\$

$= 320 \frac{m}{s}$

The kinetic energy of the bullet in its emerges from the block

$k=\frac{1}{2} m_1 v_1^2$

$=\frac{1}{2} \times 10^{-2} \times 320\\\\=512 \ J$

3 0

3 years ago

In the diagram, q1= -2.60*10^-9 C and

Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

$E_{net}=90.37\: N/c$

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

Let's find first E1 and E2.

The electric field equation is given by:

$|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}$

Where:

k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
q1 is the first charge
d1 is the distance from q1 to P

$|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}$

$|E_{1}|=80.84\: N/C$

And E2 will be:

$|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}$

$|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}$

$|E_{2}|=40.39\: N/C$

Finally, we need to use the Pythagoras theorem to find the magnitude of the net electric field.

$E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}$

$E_{net}=\sqrt{80.84^{2}+40.39^{2}}$

$E_{net}=90.37\: N/c$

I hope it helps you!

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3 years ago

Which three factors are used to calculate gravitational potential energy?

NeX [460]

Answer:

Height, mass, acceleration.

Explanation:

I hope it helps u dear! ^_^

7 0

1 year ago