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sashaice [31]
3 years ago
7

Can crushers help us recycle in a space efficient way which is good for saving the earth and for giving you more room in your ap

artment. Let us model the illustrated crusher as a planar mechanism that is subjected to the pushing force on the lower "L-shaped" handle, i.e. link CDE. Note that link CDE is one solid piece, pinned at D and E. Knowing that the orange horizontal member has a square peg at that slides vertically in the slot on the blue frame, that the can is centered under the pin at and that N, , , mm, mm, mm, and mm, determine the force magnitude in N exerted on the can from the mechanism. N

Engineering
1 answer:
777dan777 [17]3 years ago
6 0

Answer:

Force magnitude = 296.7 N

Explanation:

Detailed illustration is given in the attached document.

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A piston–cylinder device contains 15 kg of saturated refrigerant-134a vapor at 280kPa. A resistance heater inside the cylinder w
attashe74 [19]

Answer:

18.77 A

Explanation:

To solve this we use the energy balance equation, that is:

E_{in}-E_{out}=\Delta E_{sys}\\\\Q_{in}+W_{in}+W_{out}=\Delta U\\\\Q_{in}+W_{in}=\Delta U\\\\But\ W_{in}=Voltage(V)*Current(I)*change\ in\ time(\Delta t)=VI\Delta t,\Delta U=m(h_2-h_1)\\\\Given\ that\ m=15kg,V=110\ V,\Delta t=6\ min = (6*60\ s)=360\ s,Q_{in}=20.67\ kJ/kg*15\ kg=310.05\ kJ=310050\ J\\\\From\ table: At\ P_1=280kPa, h_1=249.71\ kJ/kg=249710\ J/kg;At\ P_2=280kPa \ and\  T_1=75^oC,P_2=319.95\ kJ/kg=319950\ J/kg\\\\Substituting:\\\\310500+(110*360*I)=15(319950-249710)\\\\

39600I=1053600-310500\\\\39600I=743100\\\\I=18.77\ A

3 0
3 years ago
............ ..........<br>​
sukhopar [10]

Answer:

...................

Explanation:

6 0
3 years ago
MODIFIED-BOTTOM-UP-CUT-ROD(p, n, c) to return not only the value but the actual solution, too. Hint: It is similar to how array
Vaselesa [24]

Answer:

b.

Matrix chain multiplication

M[i,j] = M[i,k] + M[(k+1),j] + p[i-1]*p[k]*p[j] i<=k<j

p[] = {5,10,3,12,5,50}

M[0][0] = 0,M[1][1] = 0,M[2][2] = 0,M[3][3] = 0,M[4][4] = 0,M[5][5] = 0,

M[1][2] = M[1][1]+M[2][2]+p[0]*p[1]*p[2] = 0+0+5*10*3 = 150

M[2][3] = M[3][3]+M[2][2]+p[1]*p[2]*p[3] = 0+0+10*3*12 = 360

M[3][4] = M[3][3]+M[4][4]+p[2]*p[3]*p[4] = 0+0+3*12*5 = 180

M[4][5] = M[4][4]+M[5][5]+p[3]*p[4]*p[5] = 0+0+12*5*50 = 3000

M[1][3] = min{M[1][1]+M[2][3]+p[0]*p[1]*p[3] , M[1][2]+M[3][3]+p[0]*p[2]*p[3]}

= {0 + 360 + 600 , 150+0+180} = {960,330} = 330

M[2][4] = min{M[2][2]+M[3][4]+p[1]*p[2]*p[4] , M[2][3]+M[4][4]+p[1]*p[3]*p[4]}

= {0 + 180 + 150 , 360+0+600} = {960,330} = 330

M[3][5] = min{M[3][3]+M[4][5]+p[2]*p[3]*p[5] , M[3][4]+M[5][5]+p[2]*p[4]*p[5]}

= {0 + 3000 + 1800 , 180+0+750} = {4800,930} = 930

M[1][4] = min{M[1][1] + M[2][4] +p[0]*p[1]*p[4] ,M[1][2] + M[3][4] +p[0]*p[2]*p[4] ,

M[1][3] + M[4][4] +p[0]*p[3]*p[4]}

{0+330+250 , 150+180+75 , 330+0+300} = 405

M[2][5] = min{M[2][2] + M[3][5] +p[1]*p[2]*p[5] ,M[2][3] + M[4][5] +p[1]*p[3]*p[5] ,

M[2][4] + M[5][5] +p[1]*p[4]*p[5]}

{0+930+1500 , 360+3000+6000,330+0+2500} = 2430

M[1][5] = min{M[1][1] +M[2][5]+p[0]*p[1]*p[5] , M[1][2] +M[3][5]+p[0]*p[2]*p[5],

M[1][3] +M[4][5]+p[0]*p[3]*p[5] , M[1][4] +M[5][5]+p[0]*p[4]*p[5]}

{0+2430+2500 , 150+930+750 , 330+3000+3000 , 405+0+1250} = 1655

(a)

MemoizedCutRod(p, n)

r: array(0..n) := (0 => 0, others =>MinInt)

return MemoizedCutRodAux(p, n, r)

MemoizedCutRodAux(p, n, r)

if r(n) = 0 and then n /= 0 then -- check if need to calculate a new solution

q: int := MinInt

for i in 1 .. n loop

q := max(q, p(i) + MemoizedCutRodAux(p, n-i, r))

end loop

end if

r(n) := q

end if

return r(n)

8 0
3 years ago
What is the power system?
Inessa [10]

Answer: An electric power system is a network of electrical components deployed to supply, transfer, and use electric power.

Explanation:

4 0
4 years ago
Show the bias polarities and depletion regions of an npn BJT in the normal active, saturation, and cutoff modes of operation. Dr
Troyanec [42]

Complete Question:

Show the bias polarities and depletion regions of an npn BJT in the normal active, saturation, and cutoff modes of operation. Draw the three sketches one below the other to (qualitatively) reflect the depletion widths for these biases, and the relative emitter, base, and collector doping.

Consider a BJT with a base transport factor of 1.0 and an emitter injection efficiency of 0.5.

Calculate roughly by what factor would doubling the base width of a BJT would increase, decrease, or leave unchanged the emitter injection efficiency and base transport factor? Repeat for the case of emitter doping increased 5 × =. Explain with key equations, and assume other BJT parameters remain unchanged!

Answer & Explanation:

[Find the attachments]

Step 1 :

Emitter and base, collector, and base are forward biased then BJT is in saturation region. Emitter and base is forward biased and base and collector in reverse biased then BJT is in active region.

Emitter and base, collector and base are reverse biased then BJT in cut off region.

Three sketches one below the other is shown in Figure 1.

[find the figure in attachment]

Step 2:

Value of base widths of saturation, active and cut off operated BJT are value of Base width of saturated region operated BJT is less than base width in active region operated BJT. Value of base width of active region operated BJT is less than base width in cut off region operated BJT.

Saturation region operated base width of BJT is < Active region operated base width of BJT is < Cut off region operated base width of BJT.

[For  Steps 3 4 5 6 and 7 find attachments]

8 0
4 years ago
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