A box is pushed horizontally with constant speed across a rough horizontal surface.

Analogue signals transmit information for such things as _____________.

ivann1987 [24]

Transmission of information in ANY form can be done digitally
or analoguely.

Beginning about 30 years ago, everything slowly started changing
to digital. Today, all commercial satellite communication, all optical
fiber communication, all internet communication, all computer
communication, all commercial cable communication, all commercial
television, and much of the telephone system, are all digital.

On your computer ... .pdf, .jpg, .mp3 etc. are all digital methods of
moving and storing information.

AM and FM radio are an interesting subject. They're all still analog.
They could easily be changed to all digital, and it would be a big
improvement, both for the broadcasters and for the listeners.
BUT ... every AM and FM radio that anybody has now would be
obsolete. Every single radio would either need to be replaced,
OR you'd need to add a digital decoder to every radio, like we
had to do with our TV sets a few years ago when television
suddenly became all digital. With AM and FM radios, the decoders
would be bigger, and would cost more, than most of the radios.

And that's why commercial radio broadcasting is still analog.

7 0

2 years ago

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A relaxed spring of length 0. 13 m stands vertically on the floor; its stiffness is 1180 N/m. You release a block of mass 0. 5 k

Daniel [21]

The length of spring when the block comes momentarily to rest on the compressed spring will be 0.054 m.The length of the spring by the letter x.

<h3>What is the potential energy of the spring?</h3>

The energy is stored in the spring when it is stretched or compressed by some length. It is the product of mass, gravity and distance compressed or stretched. Mathrmatically it is given by;

PE=mgh

The given data in the problem is ;

m is the mass of the block is 0.5 kg height,

h is the height is released is 0.7 m

x initial length of the spring = 0.13 m.

K is the force constant of the spring = 1180 N/m.

By the law of conservation of energy,

The potential energy of the spring gets converted

$\rm mgh=\frac{1}{2}KL^2 \\\\ \rm L= \sqrt{\frac{2mgh}{K} } \\\\ \rm L= \sqrt{\frac{2\times 0.5\times9.81 \times 0.7}{1180} \\\\ \\\\$