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.
I think it’s B open series circuit but I’m not sure
Answer: Approximately 8.0g of water
Explanation:
Answer:
5. 9GmM/(10R)
Explanation:
m is the mass of the satellite
M is the mass of the earth
W is the energy required to launch the satellite
Energy at earth surface = Potential energy (PE) + W
W = Energy at earth surface - Potential energy (PE)
But PE = 
Therefore: W = Energy at earth surface - 
Energy at earth surface (E) at an altitude of 5R = 
But 
Therefore: 
W = E - PE
Answer:
the gauge pressure at the upper face of the block is 116 Pa
Explanation:
Given the data in the question;
A cubical block of wood, 10.0 cm on a side.
height h = 1.50 cm = ( 1.50 × ( 1 / 100 ) ) m = 0.0150 m
density ρ = 790 kg/m³
Using expression for the gauged pressure;
p-p₀ = ρgh
where, p₀ is atmospheric pressure, ρ is the density of the substance, g is acceleration due to gravity and h is the depth of the fluid.
we know that, acceleration due to gravity g = 9.8 m/s²
so we substitute
p-p₀ = 790 kg/m³gh × 9.8 m/s² × 0.0150 m
= 116.13 ≈ 116 Pa
Therefore, the gauge pressure at the upper face of the block is 116 Pa
