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FromTheMoon [43]

3 years ago

11

A single fixed pulley is used to lift a load of 400N by the application of an effort of 480N in 10s through a vertical height of

5m. Determine the V.R, M.A and efficiency of the machine.

1 answer:

Allushta [10]3 years ago

4 0

Answer:

(a) the velocity ratio of the machine (V.R) = 1

(b) The mechanical advantage of the machine (M.A) = 0.833

(c) The efficiency of the machine (E) = 83.3 %

Explanation:

Given;

load lifted by the pulley, L = 400 N

effort applied in lifting the, E = 480 N

distance moved by the effort, d = 5 m

(a) the velocity ratio of the machine (V.R);

since the effort applied moved downwards through a distance of d, the load will also move upwards through an equal distance 'd'.

V.R = distance moved by effort / distance moved by the load

V.R = 5/5 = 1

(b) The mechanical advantage of the machine (M.A);

M.A = L/E

M.A = 400 / 480

M.A = 0.833

(c) The efficiency of the machine (E);

$E = \frac{M.A}{V.R} \times 100\%\\\\E = 0.833 \ \times \ 100\%\\\\ E = 83.3 \ \%$

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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

Neon is compressed from 100 kPa and 20◦C to 500 kPa in an isothermal compressor. Determine the change in the specific volume and

PIT_PIT [208]

Answer:

The specific volume is reduced in 80 per cent due to isothermal compression.

Specific enthalpy remains constant.

Explanation:

Let suppose that neon behaves ideally, the equation of state for ideal gases is:

$P\cdot V = n\cdot R_{u}\cdot T$

Where:

$P$ - Pressure, measured in kilopascals.

$V$ - Volume, measured in cubic meters.

$n$ - Molar quantity, measured in kilomoles,

$T$ - Temperature, measured in kelvins.

$R_{u}$ - Ideal gas constant, measured in $\frac{kPa\cdot m^{3}}{kmol\cdot K}$ .

On the other hand, the molar quantity ( $n$ ) and specific volume ( $\nu$ ), measured in cubic meter per kilogram, are defined as:

$n = \frac{m}{M}$ and $\nu = \frac{V}{m}$

Where:

$m$ - Mass of neon, measured in kilograms.

$M$ - Molar mass of neon, measured in kilograms per kilomoles.

After replacing in the equation of state, the resulting expression is therefore simplified in term of specific volume:

$P\cdot V = \frac{m}{M}\cdot R_{u}\cdot T$

$P\cdot \nu = \frac{R_{u}\cdot T}{M}$

Since the neon is compressed isothermally, the following relation is constructed herein:

$P_{1}\cdot \nu_{1} = P_{2}\cdot \nu_{2}$

Where:

$P_{1}$ , $P_{2}$ - Initial and final pressure, measured in kilopascals.

$\nu_{1}$ , $\nu_{2}$ - Initial and final specific volume, measured in cubic meters per kilogram.

The change in specific volume is given by the following expression:

$\frac{\nu_{2}}{\nu_{1}} = \frac{P_{1}}{P_{2}}$

Given that $P_{1} = 100\,kPa$ and $P_{2} = 500\,kPa$ , the change in specific volume is:

$\frac{\nu_{2}}{\nu_{1}} = \frac{100\,kPa}{500\,kPa}$

$\frac{\nu_{2}}{\nu_{1}} = \frac{1}{5}$

The specific volume is reduced in 80 per cent due to isothermal compression.

Under the ideal gas supposition, specific enthalpy is only function of temperature, as neon experiments an isothermal process, temperature remains constant and, hence, there is no change in specific enthalpy.

Specific enthalpy remains constant.

8 0

3 years ago

Why does my delivery date keep changing on my tesla model 3

telo118 [61]

Because i stole it lol

5 0

3 years ago

A ½-hp motor requires 120 volts. What amperage is required?

Zolol [24]

Answer:

3.11 A

Explanation:

Use the given relations between power, horsepower, voltage, and current to find the current requirement for the motor.

P = hp(745.7 W) = (1/2)(745.7 W) = 372.85 W

I = P/V = (372.85 W)/(120 V) = 3.1071 A

The amperage required is about 3.11 A.

5 0

3 years ago

Given: A graphite-moderated nuclear reactor. Heat is generated uniformly in uranium rods of 0.05m diameter at the rate of 7.5 x

sineoko [7]

Answer:

The maximum temperature at the center of the rod is found to be 517.24 °C

Explanation:

Assumptions:

1- Heat transfer is steady.

2- Heat transfer is in one dimension, due to axial symmetry.

3- Heat generation is uniform.

Now, we consider an inner imaginary cylinder of radius R inside the actual uranium rod of radius Ro. So, from steady state conditions, we know that, heat generated within the rod will be equal to the heat conducted at any point of the rod. So, from Fourier's Law, we write:

Heat Conduction Through Rod = Heat Generation

-kAdT/dr = qV

where,

k = thermal conductivity = 29.5 W/m.K

q = heat generation per unit volume = 7.5 x 10^7 W/m³

V = volume of rod = π r² l

A = area of rod = 2π r l

using these values, we get:

dT = - (q/2k)(r dr)

integrating from r = 0, where T(0) = To = Maximum center temperature, to r = Ro, where, T(Ro) = Ts = surface temperature = 120°C.

To -Ts = qr²/4k

To = Ts + qr²/4k

To = 120°C + (7.5 x 10^7 W/m³)(0.025 m)²/(4)(29.5 W/m.°C)

To = 120° C + 397.24° C

<u>To = 517.24° C</u>

5 0

4 years ago