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3 years ago

What do machines require in order to work?

Engineering

1 answer:

FrozenT [24]3 years ago

3 0

Answer:

Fuel and occasionaly electricity.

Explanation:

Nothing works without some kind of fuel.

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The collar A, having a mass of 0.75 kg is attached to a spring having a stiffness of k = 200 N/m . When rod BC rotates about the

gladu [14]

Answer:

Speed=1.633 m/s

Force= 20 N

Explanation:

Ideally, $v^{2}=\frac {ks^{2}}{m}$ hence $v=s\sqrt {\frac {k}{m}}$ where v is the speed of collar, m is the mass of collar, k is spring constant and s is the displacement.

In this case, s=100-0=100mm=0.1m since 1 m is equivalent to 1000mm

k is given as 200 N/m and mass is 0.75 Kg

Substituting the given values

$v=0.1 m\sqrt \frac {200 N/m}{0.75 Kg}=1.632993162 m/s\approx 1.633 m/s$

Therefore, <u>the speed is 1.633 m/s</u>

The sum of vertical forces is given by mg where g is acceleration due to gravity and it's value taken as $9.81 m/s^{2}$

Therefore, $F_y=0.75\times 9.81=7.3575 N\approx 7.36 N$

The sum of forces in normal direction is given by $Ma_n=Ks$ therefore

$Ma_n=200*0.1=20 N$

Therefore, <u>normal force on the rod is 20 N</u>

5 0

3 years ago

The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp

Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

$Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1$

The Lower Surface Cp is given as

$Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1$

The difference of the Cp over the airfoil is given as

$\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32$

Now the Lift Coefficient is given as

$C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192$

Now the coefficient of moment about the leading edge is given as

$C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306$

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

5 0

3 years ago

The lift on a spinning circular cylinder in a freestream with a velocity of 30 m/s and at standard sea level conditions is 6 N/m

Evgesh-ka [11]

Answer:

The circulation around the cylinder is 0.163 $\frac{m^{2} }{s}$

Explanation:

Given :

Velocity of spinning cylinder $v = 30$ $\frac{m}{s}$

Sea level density $\rho = 1.23$ $\frac{kg}{m^{3} }$

Sea level span $L = 6$ $\frac{N}{m}$

Lift per unit circulation is given by,

$L = \rho v c$

Where $c =$ circulation around cylinder

$c = \frac{L}{\rho v}$

$c = \frac{6}{1.23 \times 30}$

$c = 0.163$ $\frac{m^{2} }{s}$

Therefore, the circulation around the cylinder is 0.163 $\frac{m^{2} }{s}$

5 0

3 years ago

Technician A says test lights are great for performing simple tests. Technician B says you can use a test light to check SRS cir

adoni [48]

The technician that is correct about either testing lights for simple tests or to check SRS Circuits is; Technician A.

<h3>Which Technician is Correct?</h3>

First of all it is pertinent to note that test lights are generally small bulbs that are turned on by the voltage and current flowing through the circuit in analog circuits.

Now, the two values of voltage and current are high and sufficient to light up the bulb. However, in digital circuits, the current is very small in the order of milliamps, and as a result there is not enough power to turn on the lights.

Thus, we can conclude that Technician A is correct.

Read more about Correct Technician at; brainly.com/question/14449935

5 0

2 years ago

A binary geothermal power plant uses geothermal water at 160°C as the heat source. The cycle operates on the simple Rankine cycl

bogdanovich [222]

A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycle are to be determined

Assumptions :

1. Steady operating conditions exist.

2. Kinetic and potential energy changes are negligible.

Properties: The specific heat of geothermal water ( $c_{geo}$ [) is taken to be 4.18 kJ/kg.ºC.

Analysis (a) We need properties of isobutane, we can obtain the properties from EES.

a. Turbine

P $P_{3} = 3.25mPa = (3.25*1000) kPa\\= 3250kPa\\from the EES TABLE\\h_{3} = 761.54 kJ/kg\\s_{3} = 2.5457 kJ/kg\\P_{4} = 410kPa\\\\s_{4} = s_{3} \\h_{4s} = 470.40kJ/kg\\\\T_{4} = 179.5^{0} C\\\\h_{4} = 689.74 kJ/KG\\\\ The isentropic efficiency, n_{T} = \frac{h_{3}-h_{4} }{h_{3}- h_{4s} }$

= $=\frac{761.54-689.74}{761.54-670.40} \\=\frac{71.8}{91.14} \\=0.788$

b. Pump

$h_{1} = h_{f} @ 410kPa = 273.01kJ/kg\\v_{1} = v_{f} @ 410kPa = 0.001842 m^{3}/kgw_{p,in} = \frac{v_{1}(P_{2}-P_{1}) }{n_{p} } \\\\= \frac{0.01842(3250-410)}{0.9} \\\\ =5.81kJ/kg\\h_{2} =h_{1} + w_{p,in}\\ = 273.01+5.81\\ = 278.82 kJ/kg\\\\w_{T,out} = m^{.} (h_{3} -h_{4} )\\=(305.6)(761.54-689.74)\\=305.6(71.8)\\=21,942kW\\\\$

$W^{.} _ {P,in} = m^{.} (h_{2} -h_{1}) \\=m^{.} w_{p,in \\=305.6(5.81)\\\\=1,777kW\\W^{.} _{net} = W^{.} _{T, out} - W^{.} _{P,in} \\= 21,942-1,777\\=20,166 kW\\\\HEAT EXCHANGER\\\\Q_{in} = m^{.} _{geo} c_{geo} (T_{in-T_{out} } )\\=555.9(4.18)(160-90)\\=162.656kW\\$

c. $The thermal efficiency of the cycle n_{th} =\frac{W^{.} _{net} }{Q^{._{in} } } \\\\= \frac{20,166}{162,656} \\=0.124\\=12.4%$

7 0

3 years ago

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