Answer:
<h2>4.6 m/s²</h2>
Explanation:
The acceleration of an object given it's velocity and time taken can be found by using the formula
<h3>
</h3>
where
v is the final velocity
u is the initial velocity
t is the time taken
a is the acceleration
Since the body is from rest u = 0
From the question we have
We have the final answer as
<h3>4.6 m/s²</h3>
Hope this helps you
The correct answer is the third one: move toward the ground state. Remember please that Elements produce their spectrum when their electrons move toward the ground state. Hope this is very useful
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
<em>(1,500) · (0.966)^x</em>
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 <u>0.15dB per meter</u>.
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) = <em><u>-0.15</u> </em> <== 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 = <em>8.3%</em> <== <u>That's</u> 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.
(a) 
The change in energy of the transferred charge is given by:
where
q is the charge transferred
is the potential difference between the ground and the clouds
Here we have
So the change in energy is
(b) 7921 m/s
If the energy released is used to accelerate the car from rest, than its final kinetic energy would be
where
m = 950 kg is the mass of the car
v is the final speed of the car
Here the energy given to the car is
Therefore by re-arranging the equation, we find the final speed of the car:
