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3 years ago

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9. An elevator on a construction site is being operated at rated capacity of 6 tons, and is supported by two standard steel cabl

es, each having a cross-section area of 1.25-in.2 and an effective modulus of elasticity of 12 * 106 psi. As the elevator is descending at a constant 500 ft/min, an accident causes the top of the cable, 200 ft above the elevator, to stop suddenly. (a) Calculate the Impact Factor. (b) Estimate the maximum elongation and maximum tensile stress developed in cable. (c) Estimate the maximum elongation and maximum tensile stress developed in cable if the cable length was 100 ft. (d) Comment on the effect of changing cable length on maximum tensile stress developed in the cable.

1 answer:

aleksklad [387]3 years ago

8 0

Woah that is a longgg question

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Material with hardness of 220 Vickers is harder than material with a hardness of 180 Vickers. a)-True b)- False

PSYCHO15rus [73]

Answer:

Correct option a) True.

Explanation:

It is true since the Vickers hardness value refers to the force applied in a 136 ° diamond tip penetrator divided by the surface of the groove produced in the material, the lower the impression made on this greater the value will be end of the Vickers measurement and greater its hardness.

The equation to determine the Vickers hardness value will be:

Hv= ((1.854 × P)/(d²)) (kg/mm²)

Therefore a value of 220 Vickers refers to a harder material than another value of 180 Vickers.

6 0

3 years ago

Linus is using a calculator to multiply 5,426 and 30. He enters 5,426 x 300 by mistake. What can Linus do to correct his mistake

Alexxx [7]

Answer:

<em>Linus needs to take one of the zero out and it should be 30 instead 300.</em>

Explanation:

It is because Linus put three zero instead two zero.

4 0

3 years ago

A heating system must maintain the interior of a building at TH = 20 °C when the outside temperature is TC = 2 °C. If the rate o

nasty-shy [4]

Answer:

a) $Q_H=16.4kW$

b) $w=5.467$

c) $w=1.2345$

Explanation:

From the question we are told that:

Interior temperature $TH = 20 °C$

outside temperature is $TC = 2 °C$

Rate $H=16.4kW$

a)

Generally electric resistance heating is the heat transfer from interior building

Therefore

$Q_H=R$

$Q_H=16.4kW$

b)

COP =3

Generally the equation for coefficient of performance is mathematically given by

$COP=\frac{Q_H}{w}$

Therefore

$w=\frac{Q_H}{COP}$

$w=16.4/3$

$w=5.467$

c)

Generally the equation for coefficient of performance is mathematically given by

$COP_r=\frac{TH}{TH-TC}$

$COP_r=\frac{20+273}{(20+273)-(2+273)}$

$COP_r=16.2$

Therefore

$w=1.2345$

$w=20/16.2$

$w=1.2345$

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

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