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natta225 [31]
2 years ago
5

engineers who design battery operated devices suck as sell phones and MP3 players try to make them as efficient as possible. An

engineer test a cell phone and finds that the batteries supply 10,000 J of energy to make 5,500 J of output energy in the form of sound and light for the screen. How efficient is the phone
Physics
1 answer:
ICE Princess25 [194]2 years ago
8 0
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.
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natka813 [3]

Answer:

0.775

Explanation:

The weight of an object on a planet is equal to the gravitational force exerted by the planet on the object:

F=G\frac{Mm}{R^2}

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the object

R is the radius of the planet

For planet A, the weight of the object is

F_A=G\frac{M_Am}{R_A^2}

For planet B,

F_B=G\frac{M_Bm}{R_B^2}

We also know that the weight of the object on the two planets is the same, so

F_A = F_B

So we can write

G\frac{M_Am}{R_A^2} = G\frac{M_Bm}{R_B^2}

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G\frac{0.60 M_Bm}{R_A^2} = G\frac{M_Bm}{R_B^2}

Now we can elimanate G, MB and m from the equation, and we get

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So the ratio between the radii of the two planets is

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"Unbalanced forces" show themselves as a change in the speed
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