If the 5,500 J of sound and light is the ONLY useful output
from the phone, then the phone's efficiency is
(5,500J / 10,000J) = 0.55 = 55% .
But the test engineer forgot one little minor almost insignificant detail.
As a test engineer myself, I'd say that he needs to turn in his laptop
and soldering iron, and think about changing his career to a job for
which he may be better suited, like 8 hours a day in a highway toll-booth.
What about that little radio transmitter and receiver inside the phone,
that maintain digital RF communication with the nearest cell tower ?
Without that microscopic radio transceiver ... plus 30 or 40 apps
that are always running unless you shut them off ... the device in your
pocket is essentially a flat rock with one side that sometimes glows.
Answer:d
Explanation:
Drift velocity is given by

where
=drift velocity
I=Current
n=no of electron
Q=charge of Electron
A=cross-section
If area of cross-section decreases gradually then drift velocity will increase because drift velocity is inversely proportional to Area of cross-section
First of all we state the formula
Power=work done/time
we can rearrange this formula as well
work done=power x time
Since the SI unit of time is in seconds we change the minutes to seconds
2mins= 60x2 = 120 seconds
Using our formula (work done=power x time) we simply put in the values
work done = 4500 x 120
work done = 540,000J
<span>No, because the truck applies more pressure than the bridge can support.</span>
Answer:

Explanation:
We know that charge on electron

r= 2 nm
We know that force between two charge given

Now by putting the value


We know that mass of electron
The mass of electron

F= m a
a= Acceleration of electron
a= F/m


initial velocity given that zero ,u=0

