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: The magnitude of the current in the second wire 2.67A
Explanation:
Here is the complete question:
Two straight parallel wires are separated by 7.0 cm. There is a 2.0-A current flowing in the first wire. If the magnetic field strength is found to be zero between the two wires at a distance of 3.0 cm from the first wire, what is the magnitude of the current in the second wire?
Explanation: Please see the attachments below
vector is the answer of this blank
Answer:
The distance of the object placed on the principal axis from the concave mirror.
Explanation:
In a concave mirror, the nature of the image formed formed by the object placed in front of the mirror depends on the position of the object placed in from of the mirror. It all depends on the distance between the mirror and the object placed on the principal axis.
The closer the object is to the lens, the more larger or magnified the image formed will be. For example an object placed between the focal point and the pole of a concave produces a much larger image than an object placed beyond the centre of curvature of such mirror.
