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What is the charge of an atom that has lost one electron?

Luden [163]

Answer:

positive

An atom has a even amount of protons, electrons, and a little less neutrons. If the atom lost one electron, then it now has a positive charge.

Hope it helps!

5 0

3 years ago

7. Answer each of the following questions if the student applied a 550N force to a 100kg box on the same surface. What would be

blagie [28]

F=ma
550=100a
a=550/100
a=5.5 m/s²

5 0

3 years ago

S GP A projectile of mass m moves to the right with a speed vi (Fig. P11.51a). The projectile strikes and sticks to the end of a

Andrews [41]

What is the kinetic energy of the system after the collision?

$K_f=\frac{3}{2} \frac{m^{2}v_i^{2} }{(M+3m)}$

How this is calculated?

Given:

Initial speed= $v_i$

mass of rod=M

Let, Initial kinetic energy = $K_i$

Final kinetic energy= $K_f$

Moment of inertia =I

What is the moment of inertia?

$I=(I_p)_0+(I_{rod})_0\\I=m(\frac{d}{2})^{2} +\frac{Md^{2} }{12} \\I=\frac{(M+3m)d^{2} }{12}$

What is the angular momentum?

By conservation of angular momentum,

$L_i=L_f$

$mv_i\frac{d}{2}=\frac{(M+3m)d^{2}\omega }{12} \\\omega=\frac{6mv_i^{2} }{d(M+3m)}$

We know that, the final kinetic energy is given by,

$K_f=I\omega^{2}\\K_f=\frac{1}{2} *\frac{(M+3m)d^{2} }{12} *\frac{36m^{2}v_i^{2}}{d^{2}(M+3m)^{2}}\\ K_f=\frac{3}{2} \frac{m^{2}v_i^{2} }{(M+3m)}$

What is the kinetic energy?

In physics, the kinetic energy of an object is the energy that it possesses due to its motion.
It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.
Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes.

To know more about kinetic energy, refer:

brainly.com/question/114210

#SPJ4

8 0

1 year ago

Fiber-optic cables are used widely for internet wiring, data transmission, and surgeries. When light passes through a fiber-opti

Gwar [14]

After one meter, 3.4% of the light is gone ... either soaked up in the fiber
material or escaped from it. So only (100 - 3.4) = 96.6% of the light
remains, to go on to the next meter.

After the second meter, 96.6% of what entered it emerges from it, and
that's 96.6% of 96.6% of the original signal that entered the beginning
of the fiber.

==> After 2 meters, the intensity has dwindled to (0.966)² of its original level.
It's that exponent of ' 2 ' that corresponds to the number of meters that the light
has traveled through.

==> After 'x' meters of fiber, the remaininglight intensity is (0.966) ^x-power
of its original value.

If you shine 1,500 lumens into the front of the fiber, then after 'x' meters of
cable, you'll have
(1,500) · (0.966)^x
lumens of light remaining.

=========================================

The genius engineers in the fiber design industry would not handle it this way.
When they look up the 'attenuation' of the cable in the fiber manufacturer's
catalog, it would say "15dB per 100 meters".

What does that mean ? Break it down: 15dB in 100 meters is 0.15dB per meter.
Now, watch this:

Up at the top, the problem told us that the loss in 1 meter is 3.4% . We applied
super high mathematics to that and calculated that 96.6% remains, or 0.966.

Look at this ==> 10 log(0.966) = -0.15  <== loss per meter, in dB .

Armed with this information, the engineer ... calculating the loss in 'x' meters of
fiber cable, doesn't have to mess with raising numbers to powers. All he has to
do is say ...

-- 0.15 dB loss per meter

-- 'x' meters of cable

-- 0.15x dB of loss.

If 'x' happens to be, say, 72 meters, then the loss is (72) (0.15) = 10.8 dB .

and 10 ^ (-10.8/10) = 10 ^ -1.08 = 0.083 = 8.3% <== That's how much light
he'll have left after 72 meters, and all he had to do was a simple multiplication.

Sorry. Didn't mean to ramble on. But I do stuff like this every day.

5 0

3 years ago

Which of the following statements concerning statistical probability is not true? A. Probability involves the level of certainty

Leto [7]

The answer to this is D. Probability levels for finding statistical significance increase as statistical power decreases. The statement concerning statistical probability is not true as this is d.

I hope this helps!

8 0

3 years ago

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