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Feliz [49]
3 years ago
6

A town requiring 2.0 m3/s of drinking water has two sources, a local well with 15 g/m3 nitrate (as N) and a distant reservoir wi

th 5 g/m3 nitrate (as N). What flow rates of well and reservoir water are needed to meet the EPA drinking water standard and minimize the use of more expensive reservoir water
Physics
1 answer:
kati45 [8]3 years ago
8 0

Explanation:

Drinking water requirement in town 2.0 m^3/s of water per second

nitrate in local well  nitrate per 15 \mathrm{m}^{3} of water

nitrate in distant reservoir =5 \mathrm{~g} / \mathrm{m}^{3}

Let the flow rate of well

flow rate of reservoir =y m^{3} / s

Drinking water requirement is 45 \mathrm{ppm} or 45 \mathrm{~g} / \mathrm{m}^{3}

therefore, the total flow of drinking water

 

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A pendulum takes 10 seconds to swing through 2 complete cycles what is it’s period and frequency
Orlov [11]

its period should be the amount it takes to cycle from cycle to cycle so it would be 10 and your frequency would have to be calculated by taking 10 and dividing by 2 since that is how many cycles you have done so your frequency is 5


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7 0
3 years ago
Read 2 more answers
You have a pendulum clock made from a uniform rod of mass M and length L pivoting around one end of the rod. Its frequency is 1
drek231 [11]

The new oscillation frequency of the pendulum clock is 1.14 rad/s.

     

The given parameters;

  • <em>Mass of the pendulum, = M </em>
  • <em>Length of the pendulum, = L</em>
  • <em>Initial angular speed, </em>\omega _i<em> = 1 rad/s</em>

The moment of inertia of the rod about the end is given as;

I_i = \frac{1}{3} ML^2

The moment of inertia of the rod between the middle and the end is calculated as;

I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} =  \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}

Apply the principle of conservation of angular momentum as shown below;

I _i \omega _i = I _f \omega _f\\\\\frac{ML^2}{3} (1 \ rad/s)= \frac{7ML^2}{24} \times \omega _f\\\\\frac{24 \times ML^2}{3 \times 7 ML^2} (1 \ rad/s)= \omega _f\\\\1.14 \ rad/s = \omega _f

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.

Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129

3 0
3 years ago
Please help ASAP. In what direction do you need to apply force to move an object vertically?
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5 0
3 years ago
Denise is conducting a physics experiment to measure the acceleration of a falling object when it slows down and comes to a stop
77julia77 [94]

Answer: The answer is D. (9.8 m/s2)

Explanation:

7 0
3 years ago
A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo
Talja [164]

Answer:

T = 16.72 N

Explanation:

When the object is swung in a circular path, and in a vertical plane, there are two forces external to the object acting on it at any time: the gravity (which is always downward) and the tension in the string (which always points towards the center of the circle).

At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac =  v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

5 0
3 years ago
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