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Marina CMI [18]
3 years ago
8

Cell phones are technology that is popular throughout the world. Would this technology be of use in rural areas of the world

Physics
2 answers:
bonufazy [111]3 years ago
6 0
Unless there is service it wifi, no. To access most features on a cellphone you need wifi or service. These features can include communicating with others or watching a video. Cellphones are practically useless without internet connection.
katovenus [111]3 years ago
6 0

Answer:

Explanation:

Cell phone is not only a device to set up a call, but it also provides various features now a days which keeps us connected to the world. It even help one to learn by providing various means of applications.

This technology is mostly used in urban area as they are equipped with wifi connectivity and antennas. If rural areas are provided with such technologies and  advancement, then cell phones are well suited for such areas.

The advancement in technology in terms of mobile is vast and connects people through out the world.

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Answer:

Check body of the explanation

Explanation:

Ooook, quick theory rushdown. if you're at a depth of h in a tank of a fluid, the pressure is the sum of the atmosferic pressure (if the tank is open on top) plus a term which is the product of acceleration of gravity - about 10 ms^-^2, the density of whatever you're sinking in, and the depth at which you are. In formula, p(h) = p_0 + \rho g h, and the pressure is the same for every point of the tank at the same depth.

At this point, we can start answering!

1a. The pressure at A is - not counting atmosferic pressure - 1000 * 10 * 1 = 10^4 Pa, while in B is 1000*10*2 = 2*10^4 Pa, so it's half of it.

1b. The two points are at the same depth, so the pressure is the same - they would be even if the two cilinders weren't linked!

1c. Ditto. Same depth? same pressure!

1d. Usual equation, this time density is 800. Pressure is 800*10*2 = 1,6*10^4 Pa: Since the density is 4/5 of water, the pressure is also 4/5 of the one exerted by water

2a. The volume is simply the product, so 4m*3m*2m = 24m^3

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2c. Weight is defined as the mass of something times the acceleration due to gravity, in our case it's 1.92 *10^4 kg * 10 ms^{-2} = 1.92 * 10^5 N

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3. Yet again, \rho gh. 1000 {kg \over m^3} * 10 {N \over kg}} * 2 m = 2* 10^4 {N\over m^2} =2*10^4 Pa

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