What speeds did john j montgomerys gliders reach

A large well-mixed tank of unknown volume, open to the atmosphere initially, contains pure water. The initial height of the solu

trasher [3.6K]

Answer:

The exact time when the sample was taken is = 0.4167337 hr

Explanation:

The diagram of a sketch of the tank is shown on the first uploaded image

Let A denote the first inlet

Let B denote the second inlet

Let C denote the single outflow from the tank

From the question we are given that the diameter of A is = 1 cm = 0.01 m

Area of A is = $\frac{\pi}{4}(0.01)^{2} m^{2}$

= $7.85 *10^{-5}m^{2}$

Velocity of liquid through A = 0.2 m/s

The rate at which the liquid would flow through the first inlet in terms of volume = $\frac{Volume of Inlet }{time}$ = $Velocity * Area$ i.e is $m^{2} * \frac{m}{s} = \frac{m^{3}}{s}$

= $0.2 *7.85*10^{-5} \frac{m^{3}}{s}$

The rate at which the liquid would flow through the first inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

= $1039.8 * 0.2 * 7.85 *10^{-5} Kg/s$

= 0.016324 $\frac{Kg}{s}$

From the question the diameter of B = 2 cm = 0.02 m

Area of B = $\frac{\pi}{4} * (0.02)^{2} m^{2} = 3.14 * 10^{-4}m^{2}$

Velocity of liquid through B = 0.01 m/s

The rate at which the liquid would flow through the first inlet in terms of volume = $\frac{Volume of Inlet }{time}$ = $Velocity * Area$ i.e is $m^{2} * \frac{m}{s} = \frac{m^{3}}{s}$

= $3.14*10^{-4} *0.01 \frac{m^{3}}{s}$

The rate at which the liquid would flow through the second inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

= $1053 * 3.14*10^{-6} \frac{Kg}{s}$

= 0.00330642 $\frac{Kg}{s}$

From the question The flow rate in term of volume of the outflow at the time of measurement is given as = 0.5 L/s

And also from the question the mass of potassium chloride at the time of measurement is given as 13 g/L

So The rate at which the liquid would flow through the outflow in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

= $13\frac{g}{L} * 0.5 \frac{L}{s}$

= $\frac{6.5}{1000}\frac{Kg}{s}$ Note (1 Kg = 1000 g)

= 0.0065 kg/s

Considering potassium chloride

Let denote the rate at which liquid flows in terms of mass as as $\frac{dm}{dt}$ i.e change in mass with respect to time hence

Input(in terms of mass flow ) - output(in terms of mass flow ) = Accumulation in the Tank(in terms of mass flow )

(0.016324 + 0.00330642) - 0.0065 = $\frac{dm}{dt}$

$\int\limits {\frac{dm}{dt} } \, dx =\int\limits {0.01313122} \, dx$

=> 0.01313122 t = (m - $m_{o}$ )

From the question (m - $m_{o}$ ) is given as = 19.7 Kg

Hence the time when the sample was taken is given as

0.01313122 t = 19.7 Kg

=> t = 1500.2414 sec

t = .4167337 hours (1 hour = 3600 seconds)

5 0

4 years ago

As shown, a load of mass 10 kg is situated on a piston of diameter D1 = 140 mm. The piston rides on a reservoir of oil of depth

telo118 [61]

Answer:

165 mm

Explanation:

The mass on the piston will apply a pressure on the oil. This is:

p = f / A

The force is the weight of the mass

f = m * a

Where a in the acceleration of gravity

A is the area of the piston

A = π/4 * D1^2

Then:

p = m * a / (π/4 * D1^2)

The height the oil will raise is the heignt of a colum that would create that same pressure at its base:

p = f / A

The weight of the column is:

f = m * a

The mass of the column is its volume multiplied by its specific gravity

m = V * S

The volume is the base are by the height

V = A * h

Then:

p = A * h * S * a / A

We cancel the areas:

p = h * S * a

Now we equate the pressures form the piston and the pil column:

m * a / (π/4 * D1^2) = h * S * a

We simplify the acceleration of gravity

m / (π/4 * D1^2) = h * S

Rearranging:

h = m / (π/4 * D1^2 * S)

Now, h is the heigth above the interface between the piston and the oil, this is at h1 = 42 mm. The total height is

h2 = h + h1

h2 = h1 + m / (π/4 * D1^2 * S)

h2 = 0.042 + 10 / (π/4 * 0.14^2 * 0.8) = 0.165 m = 165 mm

7 0

3 years ago

Define waves as it applies to electromagnetic fields

julsineya [31]

Waves in the electric and magnetic fields are known as electromagnetic waves. You must first understand what a field is, which is just a technique of giving each square inch of space a numerical value. You may see that as a temperature field, for instance, when you look at the weather predictions and they mention the temperature in several locations. Every location on Earth has a unique temperature that can be quantified. Everywhere on Earth has its own wind velocity, which is another form of field. This field differs somewhat from the temperature field in that the wind velocity has both a direction and a magnitude, whereas the temperature just has a magnitude (how hot it is). A vector is a quantity that has both magnitude and direction, hence a field that contains vectors at every location is referred to as a vector field. Vector fields include the magnetic and electric fields. We may examine what would happen if we placed a charged particle at any given position in space. If the charged particle were to accelerate, we would state that the electric field there is the direction in which the particle is moving. In general, positively charged particles will move in the electric field's direction, whereas negatively charged particles will move in the opposite way. Because it is a vector field, the magnetic field exhibits comparable behavior. We discovered in the 19th century that the same interaction, electromagnetism, really produces both electric and magnetic fields. Like an electromagnet, a changing electric field will produce a magnetic field, and a changing magnetic field will induce an electric field (like in a generator). If your system is configured properly, you may have an electric field that fluctuates, which in turn produces a magnetic field, which in turn induces another electric field, which in turn generates another magnetic field, and so on indefinitely. At the speed of light, this oscillation between a strong magnetic field and strong electric field spreads out indefinitely. In reality, light is an electromagnetic wave—an oscillation in the electromagnetic fields. An electric or magnetic field may exist without a medium since they exist in a vacuum, which implies that waves in these fields don't require a medium like sound to flow through.

5 0

2 years ago

Can a 1½ " conduit, with a total area of 2.04 square inches, be filled with wires that total 0.93 square inches if the maximum f

Papessa [141]

Answer:

it is not possible to place the wires in the condui

Explanation:

given data

total area = 2.04 square inches

wires total area = 0.93 square inches

maximum fill conduit = 40%

to find out

Can it is possible place wire in conduit conduit

solution

we know maximum fill is 40%

so here first we get total area of conduit that will be

total area of conduit = 40% × 2.04

total area of conduit = 0.816 square inches

but this area is less than required area of wire that is 0.93 square inches

so we can say it is not possible to place the wires in the conduit

4 0

3 years ago

Roads in rural areas are _______.

Roman55 [17]

Answer:

Explanation:

Mountain roads often zigzag across a mountain with a series of sharp turns called. switchbacks.

6 0

3 years ago

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