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Genrish500 [490]

3 years ago

14

6. At a construction site, cement, sand, and gravel are used to make concrete. The ratio of cement to sand to gravel is 1 to 2.4

to 3.6. If a 150-lb bag of sand is used, how much cement and gravel must be used?

1 answer:

S_A_V [24]3 years ago

7 0

Answer:

Mass of cement used is 62.5 lb

Mass of gravel used is 225 lb

Explanation:

The ratio given here is cement to sand to gravel = 1 : 2.4 : 3.6

So, for 150 lb of sand

C : S : G = 1 : 2.4 : 3.6

$\frac{C}{S}=\frac{1}{2.4}\\\Rightarrow C=S\frac{1}{2.4}\\\Rightarrow C=150\frac{1}{2.4}\\\Rightarrow C=62.5\ lb$

Mass of cement used is 62.5 lb

$\frac{S}{G}=\frac{2.4}{3.6}\\\Rightarrow G=S\frac{3.6}{2.4}\\\Rightarrow C=150\frac{3.6}{2.4}\\\Rightarrow C=225\ lb$

Mass of gravel used is 225 lb

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7 0

3 years ago

A 50 (ohm) lossless transmission line is terminated in a load with impedance Z= (30-j50) ohm. the wavelength is 8cm. Determine:

Serjik [45]

Answer:

Reflection Coefficient = $0.57e^{-i79.8}$

SWR=3.65

Position of $V_{max} =3.11cm$

position of $i_{max} =1.11cm$

Explanation:

To determine the above answers, let outline the useful formulas

refection coefficient $p=\frac{terminalimpednce -characteristics impedance }{terminalimpednce +characteristics impedance } \\$ .

where terminal impednce = (30-i50)Ω

characteristics impedance= 50Ω

Secondly, the Standing Wave Ratio, $SWR=\frac{1+/p/}{1-/p/}$

Now let us substitute values and solve,

a. $p=\frac{terminalimpednce -characteristics impedance }{terminalimpednce +characteristics impedance } \\$

$p=\frac{(30-i50)-50}{(30-i50)-50} \\$

$p=\frac{-20-i50}{80-i50} \\$

multiplying the numerator and denominator by the conjugate of the denominator. we have

$p=\frac{-20-i50}{80-i50}*\frac{80+i50}{80+i50}\\$

by carrying out careful operation, we arrived at

$p=\frac{900-i5000}{8900} \\p=0.1011-i0.56179\\$

To express in polar form i.e $re^{i alpha}$

$r=\sqrt{0.1011^{2}+0.56179^{2}} \\r=0.57\\$

to get the angle

alpha= $tan^{-1} \frac{0.56179}{0.1011} \\alpha=-79.8\\$

hence the Reflection Coefficient,<em>p</em> = $0.57e^{-i79.8}$

b. we now determine the Standing Wave Ratio, $SWR=\frac{1+/p/}{1-/p/}$

$swr=\frac{1+0.57}{1-0.57} =3.65\\$

c. to determine the position of the maximum voltage nearest to the load,

we use the equation

$Position of V_{max}=\frac{\alpha λ}{4\pi}+\frac{λ}{2}\\$

were $λ$ is the wavelength of 8cm

lets convert α to rad by multiplying by π/180

$Position of V_{max}=\frac{-79.8 *8cm*\pi}{4\pi*180 } +\frac{8cm}{2} \\$

$Position of V_{max}=-0.89+4.0=3.11cm\\$ .

d. also were we have minimum voltage,there the maximum current will exist, to find this position nearest to the load

$Position of v_{min}=Position of V_{max}-\frac{λ}{4} \\\\Position of v_{min}=3.11cm-\frac{8cm}{4}=1.11cm$ .

since the voltage minimum occure at 1.11cm. we can conclude that the current maximum also occur at this point i.e 1.11cm

3 0

3 years ago

A 15-ft beam weighing 570 lb is lowered by means of two cables unwinding from overhead cranes. As the beam approaches the ground

7nadin3 [17]

Answer:

I. Tension (cable A) ≈ 6939 lbf

II. Tension (cable B) ≈ 17199 lbf

Explanation:

Let's begin by listing out the data that we were given:

mass of beam (m) = 570 lb, deceleration (cable A) = -20 ft/s², deceleration (cable B) = -2 ft/s²,

g = 32.17405 ft/s²

The tension on an object is given by the product of mass of the object by gravitational force plus/minus the product of mass by acceleration.

Mathematically represented thus:

T = mg + ma

where:

T = tension, m = mass, g = gravitational force,

a = acceleration

I. For Cable A, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-20)]

T = 18339.2085 - 11400 = 6939.2085

T ≈ 6939 lbf

II. For Cable B, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-2)]

T = 18339.2085 - 1140 = 17199.2085

T ≈ 17199 lbf

4 0

4 years ago

You have a motor such that if you give it 12 Volt, it will eventually reach a steady state speed of 200 rad/s. If it starts from

Aleksandr [31]

Answer:

a) $\frac{Ws}{Es} = \frac{200}{1+1.2s}$

b) attached below

c) type zero system

d) k > $\frac{g}{200}$

e) The gain K increases above % error as the steady state speed increases

Explanation:

Given data:

Motor voltage = 12 v

steady state speed = 200 rad/s

time taken to reach 63.2% = 1.2 seconds

<u>a) The transfer function of the motor from voltage to speed</u>

let ; $\frac{K1}{1+St}$ be the transfer function of a motor

when i/p = 12v then steady state speed ( k1 ) = 200 rad/s , St ( time constant ) = 1.2 sec

hence the transfer function of the motor from voltage to speed

= $\frac{Ws}{Es} = \frac{200}{1+1.2s}$

<u>b) draw the block diagram of the system with plant controller and the feedback path </u>

attached below is the remaining part of the detailed solution

c) The system is a type-zero system because the pole at the origin is zero

d) ) k > $\frac{g}{200}$

7 0

3 years ago

A rectangular block of material with shear modulus G= 620 MPa is fixed to rigid plates at its top and bottom surfaces. Thelower

PIT_PIT [208]

Answer:

γ $_{xy}$ =0.01, P=248 kN

Explanation:

Given Data:

displacement = 2mm ;

height = 200mm ;

l = 400mm ;

w = 100 ;

G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²

a. Average Shear Strain:

The average shear strain can be determined by dividing the total displacement of plate by height

γ $_{xy}$ = displacement / total height

= 2/200 = 0.01

b. Force P on upper plate:

Now, as we know that force per unit area equals to stress

τ = P/A

Also, τ = Gγ $_{xy}$

By comapring both equations, we get

P/A = Gγ $_{xy}$ ------------ eq(1)

First we need to calculate total area,

A = l*w = 400 * 100= 4*10^4mm²

By putting the values in equation 1, we get

P/40000 = 620 * 0.01

P = 248000 N or 2.48 *10^5 N or 248 kN

6 0

3 years ago