Describe and compare the characteristics of (a) proportional control, (b) proportional plus integral control, (c) proportional plus integral plus derivative control.

Answer:

movement in clockwise direction.

Explanation:

The following parameters or information are given from the question above, they are:

[1]. There are two identical keys, [2]. two out of the five keys are attached to 6cm from the pivot, [3]. three keys out of the five keys on the right are attached 5 cm.

Therefore, considering the moment of force, the two keys on the left = 2 × 6 = 12.

Also, considering the moment of force, the 3 keys on the right = 3 × 5 = 15.

Therefore, we have more weight on the right keys. So, in order to balance the force there must be movement in clockwise direction.

Answer:

A) σ_y = 79096 lb/in² = 79.1 ksi

B) strain-hardening exponent = 0.102

(strength coefficient = 137838.78 lb/in²

Explanation:

A) Formula for volume is;

V = πd²h/4

We are given;

height 2.0 in and diameter 1.5 in

Thus;

V = (π × 1.5² × 2)/4

V = 3.53 in³

Area is;

A = πd²/4

A = (π × 1.5²)/4

A = 1.77 in²

Yield strength is gotten from the formula;

σ_y = Force/Area

We are given load = 140,000 lb

Thus;

σ_y = 140000/1.77

σ_y = 79096 lb/in²

B) We are given

modulus of elasticity: E = 30 × 10^(6) lb/in²

Formula for strain is;

ε = σ_y/E

ε = 79096/(30 × 10^(6))

ε = 0.00264

The metal yields (0.2% offset), thus;

strain offsets = 0.00264 + 0.002

strain offsets: ε1 = 0.00464

Thus;

(h_i - h_o)/h_o = 0.00464

(h_i/h_o) - 1 = 0.00464

(h_i/h_o) = 1.00464

h_i = h_o(1.00464)

h_o = 2 in

Thus; h_i = 2(1.00464) = 2.00928 in

Area = Volume/height = 3.53/2.00928 = 1.757 in²

True stress is;

σ = force/area = 140000/1.757

σ1 = 79681.27 lb/in²

At a load of 260,000 lb, the height has been reduced to 1.6 in. Thus;

Area = 3.53/1.6 = 2.206 in²

True stress is;

σ2 = 260000/2.206

σ2 = 117860.38 lb/in²

True strain;

ε2 = In(2/1.6)

ε2 = 0.223

From flow curve;

σ = kεⁿ

Thus;

σ1 = k(ε1)ⁿ

79681.27 = k(0.00464ⁿ) - - - (eq 1)

Also for σ2 = k(ε2)ⁿ;

117860.38 = k(0.223ⁿ) - - - - - (eq 2)

From eq 1,

k = 79681.27/0.00464ⁿ

Putting this for k in eq2 to get;

117860.38 = (0.223ⁿ) × 79681.27/0.00464ⁿ

117860.38/79681.27 = 0.223ⁿ/0.00464ⁿ

Solving for n, we have ≈ 0.102

Thus,K is;

k = 79681.27/0.00464^(0.102)

k = 137838.78 lb/in²

Answer: B is correct

Explanation:

the process in which dairies make cheese by separating a milk's solids from its liquid is called. curdling.

Answer:

a) 159.07 MPa

b) 10.45 MPa

c) 79.535 MPa

Explanation:

Given data :

length of cantilever beam = 1.5m

outer width and height = 100 mm

wall thickness = 8mm

uniform load carried by beam  along entire length= 6.5 kN/m

concentrated force at free end = 4kN

first we  determine these values :

Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m

Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N

A) determine max bending stress

б = [tex]\frac{MC}{I}[/tex]  =  [tex]\frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}[/tex]  =  159.07 MPa

B) Determine max transverse shear stress

attached below

   ζ = 10.45 MPa

C) Determine max shear stress in the beam

This occurs at the top of the beam or at the centroidal axis

hence max stress in the beam =  159.07 / 2 = 79.535 MPa  

attached below is the remaining solution

Answer:

Hello some parts of your question is missing below is the missing part

Convection coefficient = 11 w/m^2. °c

answer : 44.83 watts

Explanation:

Given data :

surface emissivity ( ε )= 0.95

head ( sphere) diameter( D )  = 0.25 m

Temperature of sphere( T )  = 35° C

Temperature of surrounding ( T∞ )  = 25°C

Temperature of surrounding surface ( Ts ) = 15°C

б  = ( 5.67 * 10^-8 )

Determine the total rate of heat loss

First we calculate the surface area of the sphere

As = [tex]\pi D^{2}[/tex]  

= [tex]\pi * 0.25^2[/tex] =  0.2 m^2

next we calculate heat loss due to radiation

Qrad = ε * б * As( [tex]T^{4} - T^{4} _{s}[/tex] )  ---- ( 1 )

where ;

ε = 0.95

б = ( 5.67 * 10^-8 )

As = 0.2 m^2

T = 35 + 273 = 308 k

Ts = 15 + 273 = 288 k

input values into equation 1

Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )

         = 22.83  watts

Qrad ( heat loss due to radiation ) = 22.83 watts

calculate the heat loss due to convection

Qconv = h* As ( ΔT )

           = 11*0.2 ( 35 -25 )  = 22 watts

Hence total rate of heat loss

=  22 + 22.83

= 44.83 watts

Answer:

8*10000+3*1000+1*00+2*10+2

Explanation:

Answer:

- Four (4) flip-flop values will complemented

- one (1) flip-flop value will complemented

Explanation:

To find how many flip flop number of bits complemented, we just need to figure out what the next count in the sequence is and find how many bits have changed.

taking a look at the a) 00110111

we need to just 1 to the value,

so

00110111 +  0000001  = 00111000        

So here, only the first four bits are complemented.

Therefore Four (4) flip-flop values will complemented

Next

b) 01010110

we also add 1 to the value

01010110  + 00000001  = 01010111

only the first bit is complemented.

Therefore one (1) flip-flop value will complemented

they're essentially the same thing so i'd say yes

Answer:

The answer is below

Explanation:

Given that:

Volume of carbide ([tex]V_{C}[/tex]) = 76% = 0.76, Volume of Nickel ([tex]V_{M}[/tex]) = 100% - 76% = 24% = 0.24, thermal conductivities of carbide ([tex]E_{C}[/tex]) = 30 W/m-K and thermal conductivities of meta. ([tex]E_M[/tex]) = 67 W/m-K

a) The maximum thermal conductivity is given by:

Max = [tex]E_CV_c+E_mV_m=(0.76*30) + (0.24*67) = 38.88\ W/mK[/tex]

b) The minimum thermal conductivity is given by:

Min = [tex]\frac{E_ME_C}{E_MV_C+E_CV_M}=\frac{30*67}{(0.76*67)+(0.24*30)}=34.58\ W/mK[/tex]

Answer: the size of memory needed for the CD recording is 28.7 MB

Explanation:

so in the case of stereo, the bitrate is;

⇒ 26 × 60 × 44.1 × 10³ × 2

=   137592 × 10³  

for 10 bit

⇒ 137592 × 10³ × 10

= 1375920 × 10³ bits

now divide by 8 (convert to bytes)

⇒ (1375920 × 10³) / 8

= 171,990,000 BYTE

divide by 1000 (convert to kilobytes)

= 171,990,000 / 1000

= 171,990 KILOBYTES

now Given that, the compression ratio is 6      

so  

171,990 / 6

= 28665 KB

we know that. 1 MB = 1000 KB

x MB = 28665 KB

x MB = 28665 / 1000

⇒ 28.665 MB ≈ 28.7 MB

Therefore the size of memory needed for the CD recording is 28.7 MB