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3 years ago

6

. Each valve A, B, and C, when open, releases water into a tank at its own constant rate. With all three valves open, the tank f

ills in 1 hour, with only valves A and C open it will take 1.5 hours, and with only valves B and C open it takes 2 hours. How many minutes will it take to fill the tank with only valves A and B open?

1 answer:

olasank [31]3 years ago

7 0

Answer:

Time taken by A and B is 1.2 hr.

Explanation:

Given that

Time taken by tank when all(A+B+C) are open = 1 hr

Time taken by tank when A+C are open = 1.5 hr

Time taken by tank when B+C are open = 2 hr

If we treat as filling of tank is a work then

Work = time x rate

Lets take work is 1 unit

1 = 1(1/a+1/b+1/c) ---------1

1 = 1.5(1/a+1/c) ----------2

1 = 2(1/b+1/c) --------3

From equation 1 and 3

1=1(1/a+1/2)

a=2

Form equation 2

1 = 1.5(1/2+1/c)

c=6

From equation 3

1 = 2(1/b+1/6)

b=3

So time taken by

A is alone to fill tank is 2 hr

B is alone to fill tank is 3 hr

C is alone to fill tank is 6 hr

So $time\ taken\ by\ A\ and\ B =\dfrac{2\times 3}{2+3} hr$

Time taken by A and B is 1.2 hr.

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If the area of a concrete slap weighing 300n is 400cm2 ,how much pressure can it apply when kept on the ground

aleksklad [387]

Answer:

7500 N/m²

Explanation:

From the question given above, the following data were obtained:

Force (F) = 300 N

Area (A) = 400 cm²

Pressure (P) =?

Next, we shall convert 400 cm² to m². This can be obtained as follow:

1×10⁴ cm² = 1 m²

Therefore,

400 cm² = 400 cm² × 1 m² / 1×10⁴ cm²

400 cm² = 0.04 m²

Finally, we shall determine the pressure. This can be obtained as follow:

Force (F) = 300 N

Area (A) = 0.04 m²

Pressure (P) =?

P = F/A

P = 300 / 0.04

P = 7500 N/m²

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3 years ago

Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude

labwork [276]

Answer: $\tau=1.03\times 10^{-4}\ N-m$

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

$\tau=NIAB\ \sin\theta$

Let us assume that, $A=0.0008\ m^2$

$\theta$ is the angle between normal and the magnetic field, $\theta=90^{\circ}-30^{\circ}=60^{\circ}$

Torque is given by :

$\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m$

So, the net torque about the vertical axis is $1.03\times 10^{-4}\ N-m$ . Hence, this is the required solution.

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3 years ago

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3 years ago

Read 2 more answers

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be:

$\displaystyle q(t) = q_0 \, e^{-t / \tau}$ .

<h3>a)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{8}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9$ .

$t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8)$ .

<h3>b)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{7}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9$ .

$t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7)$ .

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3 years ago

Animals have different types of respiratory systems depending on the environment of their habitat. Fish live in the water and us

VMariaS [17]

Answer:

1. Both are respiratory organs found in vertebrates.

2. Both enable the inhalation of oxygen and exhalation of carbon dioxide in the bodies of their host.

3. Respiratory membranes present in both make the exchange of gases easier.

4. Blood vessels found around them enable the breathing process.

Explanation:

Gills are respiratory organs which can be found in animals that live in water. These include fishes, amphibians like frogs, anthropods like insects, and annelids like worms. Gills are surrounded by respiratory membranes known as gill lamellae which have a large surface area that makes breathing easier. Blood vessels also are found around gills.

The lung is also a respiratory organ found in humans. It runs from a windpipe known as the trachea and branches into bronchi which contain the bronchioles. Alveoli ducts contain alveoli sacs and then alveoli which are membranes that enable the breathing process.

Both are similar because they are respiratory organs found in animals with backbones. They are surrounded by blood vessels and membranes that make the breathing process easier.

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3 years ago