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SOVA2 [1]
3 years ago
5

A fluid in an aquifer is 23.6 m above a reference datum, the fluid pressure (in gage pressure) is 4390 n/m2 and the flow velocit

y is 7.22 x 10-4 m/s. the fluid density is 0.999 x 103 kg/m3. using the bernoulli relationship (including velocity), calculate the (a) total energy per unit mass, and (b) total energy per unit weight. report your answer in units of meters and seconds. assume that g = 9.8 m/s2.
Physics
1 answer:
Phoenix [80]3 years ago
7 0

As per Bernuolli's Theorem total energy per unit mass is given as

\frac{P}{\rho} + \frac{1}{2}v^2 + gH = E

now from above equation

P = 4390 N/m^2

\rho = 0.999 * 10^3

v = 7.22 * 10^{-4} m/s

H = 23.6 m

now by above equation

\frac{4390}{0.999*10^3} + \frac{1}{2}*(7.22*10^{-4})^2 + 9.8*23.6 = E

E = 235.7 J/kg

Part B)

Now energy per unit weight

U = \frac{E}{g}

U = \frac{235.7}{9.8}

U = 24 m

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3 years ago
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A ball is dropped from an aircraft flying at an altitude of 8,848 meters assuming gravity is 9.8m/s what is the total amount of
Dafna11 [192]
In this question, you're determining the time (t) taken for an object to fall from a distance (d).

The equation to represent this is:
Time equals the square root of 2 times the distance divided by the gravitational force of earth.
In equation from it looks like this (there isn't an icon to represent square root so just pretend like there's a square root there):
t = 2d/g (square-rooted)

d = 8,848m and g = 9.8m/s

Now plug in the information we have:
t = 2 x 8,848m/9.8m/s (square-rooted)

The first step is to multiply 2 times 8,848m:
t = 17,696m/9.8m/s (square-rooted)

Now divide 9.8m/s by 17,696m (note that the two m's (meters) cancels out leaving you with only s (seconds):
t = 1805.72s (square-rooted)

Now for the last step, find the square root of the remaining number:
t = 42.5s

So the time it takes the ball to drop from the height (distance) of 8,848 meters, and falling with the gravitational pull of 9.8 meters per second is 42.5 seconds.

I hope this helps :)

7 0
3 years ago
Two bicycle tires are set rolling with the same initial speed of 4.0 m/s along a long, straight road, and the distance each trav
umka2103 [35]

Answer:

The coefficient of rolling friction will be "0.011".

Explanation:

The given values are:

Initial speed,

v_i = 4.0 \ m/s

then,

v_f=\frac{4.0}{2}

    =2.0 \ m/s

Distance,

s = 18.2 m

The acceleration of a bicycle will be:

⇒ a=\frac{v_f^2-v_i^2}{2s}

On substituting the given values, we get

⇒    =\frac{(2.0)^2-(4.0)^2}{2\times 18.2}

⇒    =\frac{4-8}{37}

⇒    =\frac{-4}{37}

⇒    =0.108 \ m/s^2

As we know,

⇒  f=ma

and,

⇒  \mu_rmg=ma

⇒       \mu_r=\frac{a}{g}

On substituting the values, we get

⇒       =\frac{0.108}{9.8}

⇒       =0.011

7 0
3 years ago
A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material
Llana [10]

Answer:

a) C = 40.138\,pF, b) q = 16.056\,nC, c) U = 3.212\,\mu J

Explanation:

a) The capacitance of two parallel plates capacitor with dielectric is given by the following expression:

C = K\cdot \epsilon_{o}\cdot \frac{A}{d}

Where:

K - Dielectric constant.

\epsilon_{o} - Vaccum permitivity.

A - Plate area.

d - Distance between plates.

Hence, the capacitance of the system is:

C = (4.00)\cdot (8.854\times 10^{-12}\,\frac{F}{m} )\cdot \left(\frac{17\times 10^{-4}\,m^{2}}{0.150\times 10^{-2}\,m}\right)

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b) The charge can be found by using the definition of capacitance:

q = C\cdot V_{batt}

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q = 1.606\times 10^{-8}\,C

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c) The energy stored in the charged capacitor is:

U=\frac{1}{2}\cdot Q\cdot V_{batt}

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3 0
2 years ago
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Answer:

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F*t= I

F= I/t

F= 20 N (c)

5 0
2 years ago
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