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3 years ago

If the depth of water in a well is 10m, what is the pressure exerted by it the bottom of the well ? ( Use g = 10 m/s2)

Physics

1 answer:

Tasya [4]3 years ago

7 0

Answer:

The precise answer depends on the density and therefore the temperature of the water, but we can obtain a reasonable approximation by assuming that the density of the water is 1000 kilograms per cubic meter (kg/m³).

Since the depth of the water in the well is 10 m, the volume of water directly above an area A of a square meters (m²) at the bottom of the well is 10×a m³.

Since the density of the water is 1,000 kg/m³, the mass of water directly above area A is (1,000 kg/m³) × (10×a m³) = (1000×10×a kg) = 10,000×a kg.

Since g = 9.8 m/s², the force of gravity acting on the water directly above area A is (9.8 m/s²) × (10,000×a kg) = 9.8×10,000×a N (newtons) = 98,000×a N.

So the pressure of water acting on area A is (98,000×a N)/(a m²) = (98,000×a)/a N/m² = 98,000 pascals (pa). And since A could be any given area at the bottom of the well, this is the pressure at any point at the bottom of the well.

So the pressure at the bottom of the well is 98,000 pascals (or 98,000/101,325 standard atmospheres = 560/579 atmospheres ~ 0.967 standard atmospheres).

Please comment below if you have any questions.

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Which of the following usually occurs with a short circuit? a. All parts of the circuit will begin to carry higher amounts of cu

fgiga [73]

Answer:

Hi,

The correct answer option is; D. Most of the current will flow through one part of the circuit.

Explanation:

A short circuit is a low resistance path in an electric connection between two conductors supplying current in a circuit.

It happens when excess amounts of current flow in the power source through a 'short path'.

Short circuits occur at very high temperatures which is of course caused by the heat produced during dissipation.

An example of application of short circuit is arc welding, where heating is achieved through short circuit.

6 0

3 years ago

How much force is required to accelerate a 2 kg mass at 3 m/s2

swat32

Force = (mass) x (acceleration) Newton's second law of motion.

Force = (2 kg) x (3 m/s²) = 6 newtons.

3 0

3 years ago

Mr. Galonski loves to use the Electromagnetic Spectrum. Create a scenario in which Mr. Galonski is using waves from the electrom

Anestetic [448]

Answer:

The Scenario:

On a normal Sunday afternoon Mr. Golanski is sitting in his living room reading his book. He decides after a while to turn on the Television to see what’s on the news, (Mr. Golanski is using Radio waves when he turns his television on by signaling the TV from his remote control). After a few hours Mr. Golanski decides it’s time to have dinner. He heats up a quick meal in his microwave because he doesn’t have the patience for cooking. (He is using microwave radiation to heat his food because water molecules in food absorb the radiation). He sits down for his meal, and halfway through he starts to choke! In a panicked frenzy he runs to his bathroom to try and dislodge the obstacle from his throat. By doing so he switched on the fluorescent lights in his bathroom exposing himself to small amounts of ultraviolet radiation. (Fluorescent lights absorb UV radiation and transmit visible light along with small amounts of UV light). Unable to dislodge the obstacle from his throat Mr. Golanski seeks help from his neighbor who drives him straight to the ER. To treat him properly the physicians opt for a fluoroscopy to examine Mr. Golanski’s esophageal tract. (Thus he is making use of X-ray imaging to obtain a visual of his internal esophageal structure to check for the obstruction). Once treated and discharged from the hospital Mr. Golanski returns home grateful to have survived this ordeal with minimum damage.

Explanation:

The Electromagnetic Spectrum is the range of frequencies and wavelengths for different light waves. They range with increasing frequency from Radio waves, to Microwaves, to Infrared waves, to Visible waves, to Ultraviolet waves, to Infrared waves, to X-rays, and to Gamma rays. Several of which we use in our daily lives such as Radio waves when operating our television or using our cellular phones. We also use microwaves to heat our food or for communication with satellites. We are also exposed to natural Ultraviolet radiation from the sun; however, we can also get exposed to other forms such as from certain types of light bulbs. We see visible light in the form of all the colors we can detect around us. We make use of x-rays for imaging techniques widely used in medicine for diagnostics, as well as Infrared waves in our home security systems. The electromagnetic spectrum is always used as a part of our everyday life.

7 0

3 years ago

2 pts

const2013 [10]

Answer:

I think the answer 1

Explanation:

im probably wrong too i dont know

5 0

3 years ago

While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)

ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration= $2.40m/s^{2}$ , c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = $2.40m/s^{2}$

time = 12.0 s

initial position = 0

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

$x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}$

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

$x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}$

after cancelling the pertinent units, I get that my answer will be given in meters. So I get:

$x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}$

which solves to:

$x=172.8m$

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

$a=\frac{V_{f}-V_{0} }{t}$