A rectangular block of 1m by 0.6m by 0.4m floats in water with 1/5th of its volume being out of water. Find the weight of the block.

Answers

Answer 1

Answer:

Weight of block is 191.424 Kg

Explanation:

The volume of rectangular block = [tex]1*0.6*0.4 = 0.24[/tex] cubic meter

1/5th of its volume being out of water which means water of volume nearly 4/5 th of the volume of rectangular block is replaced

Volume of replaced water = [tex]\frac{4}{5} * 0.24 = 0.192[/tex] cubic meter

Weight of replaced water = weight of rectangular block = [tex]0.192 * 997[/tex] Kg/M3

= 191.424 Kg


Related Questions

1)Saturated steam at 1.20bar (absolute)is condensed on the outside ofahorizontal steel pipe with an inside and outside diameter of 0.620 inches and 0.750 inches, respectively. Cooling water enters the tubes at 60.0°F and leaves at 75.0°F at a velocity of 6.00ft/s. (HINT: You may assume laminar condensate flow.You many also assume that the mean bulk temperature of the cooling water is equal to the wall temperature on the outside of the pipe, T".You may also neglect the viscosity correction in your calculations.)a)What are the inside

Answers

Answer:

hi = 7026.8  W/m^2.k

Explanation:

Given data :

pressure of saturated steam = 1.2 bar

Horizontal steel pipe : inside diameter = 0.620 , outside diameter = 0.750 inches

temperature of water at entry = 60°F

temperature of water at exit = 75°F

velocity of water = 6 ft/s

Calculate the Inside convective heat transfer coefficient ( hi )

mean temperature ( Tm ) = 60 + 75 / 2 = 67.5°F ≈ 292.877 K ≈ 19.727°C

next : find the properties of water at this temperature ( 19.727°C )

thermal conductivity = 0.598  w/m.k

density = 1000 kg/m^3

specific heat ( Cp ) = 4.18 KJ/kg.k

viscosity = 0.001 pa.s

velocity of water = 6 ft/s ≈ 1.8288 m/s

∴ Re ( Reynolds number ) = 28712.16

and Prandtl number ( Pr ) = (4180 * 0.001) / 0.598  = 6.989

finally to determine the inside convective heat transfer coefficient we will apply the Dittos - Bolter equation

hi = 7026.8 w/m^2.k

attached below is the remaining solution

If it is struck by a rigid block having a weight of 550 lblb and traveling at 2 ft/sft/s , determine the maximum stress in the cylinder. Neglect the mass of the cylinder. Express your answer to three significant figures and include appropriate units.

Answers

This question is incomplete, The missing image is uploaded along this answer below;

Answer:

the maximum stress in the cylinder is 3.23 ksi

 

Explanation:

Given the data in the question and the diagram below;

First we determine the initial Kinetic Energy;

T = [tex]\frac{1}{2}[/tex]mv²

we substitute

⇒ T = [tex]\frac{1}{2}[/tex] × (550/32.2) × (2)²

T = 34.16149 lb.ft

T =  ( 34.16149 × 12 ) lb.in  

T = 409.93788 lb.in

Now, the volume will be;

V = [tex]\frac{\pi }{4}[/tex]d²L

from the diagram; d = 0.5 ft and L = 1.5 ft

so we substitute

V =  [tex]\frac{\pi }{4}[/tex] × ( 0.5 × 12 in )² × ( 1.5 × 12 in )

V = 508.938 in³

So by conservation of energy;

Initial energy per unit volume = Strain energy per volume

⇒ T/V = σ²/2E

from the image; E = 6.48(10⁶) kip

so we substitute

⇒ 409.93788 / 508.938 = σ²/2[6.48(10⁶)]

508.938σ² =  5,312,794,924.8

σ² = 10,438,982.5967  

σ = √10,438,982.5967

σ = 3230.9414  

σ = 3.2309 ksi  ≈ 3.23 ksi    { three significant figures }

Therefore, the maximum stress in the cylinder is 3.23 ksi

Barries of effective
communication?

Answers

Answer: barries

Explanation:

if the output of a signal is 36% on and 64% off and repeats itself is it considered periodic
helppppp

Answers

Answer:

No,  it is not a periodic Signal

Explanation:

No,  it is not a periodic Signal

This signal is repeating itself with the fixed on and off values but the major point to note here is that is this signal repeating after a fixed length of time every time. No, such information is provided in the question and hence, this signal cannot be termed as periodic.

Which two technologies were combined to create product life cycle management (PLM) software?

CAD and a database

spreadsheets and graphics

a database and spreadsheets

CAD and spreadsheets

Answers

Answer:

CAD and a database

Explanation:

The correct answer is CAD and a database. When American Motors Corportation introduced the Jeep Cherokee, it implemented CAD to increase engineering productivity and combined that with a new communications system.

The answer is. CAD and a database

A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the other half with vapor. If it is desired that the pressure cooker not run out of liquid water for 75 min, determine the highest rate of heat transfer allowed.

Answers

Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

Volume = 4L = 0.004 m³

V[tex]_f[/tex] = V[tex]_g[/tex] = 0.002 m³

Using Table ( saturated water - pressure table);

at pressure p = 175 kPa;

v[tex]_f[/tex] = 0.001057 m³/kg

v[tex]_g[/tex] = 1.0037 m³/kg

u[tex]_f[/tex] = 486.82 kJ/kg

u[tex]_g[/tex] 2524.5 kJ/kg

h[tex]_g[/tex] = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V[tex]_f[/tex]/v[tex]_f[/tex] + V[tex]_g[/tex]/v[tex]_g[/tex]

we substitute

m₁ = 0.002/0.001057  + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v[tex]_g[/tex]

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m[tex]_{out[/tex] = m₁ - m₂

m[tex]_{out[/tex] = 1.89414  - 0.003985

m[tex]_{out[/tex] = 1.890155 kg

so, Initial internal energy will be;

U₁ = m[tex]_f[/tex]u[tex]_f[/tex] + m[tex]_g[/tex]u[tex]_g[/tex]

U₁ = (V[tex]_f[/tex]/v[tex]_f[/tex])u[tex]_f[/tex]  + (V[tex]_g[/tex]/v[tex]_g[/tex])u[tex]_g[/tex]

we substitute

U₁ = (0.002/0.001057)(486.82)  + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E[tex]_{in[/tex] -  E[tex]_{out[/tex] = ΔE[tex]_{sys[/tex]

QΔt - m[tex]_{out[/tex]h[tex]_{out[/tex] = m₂u₂ - U₁

QΔt - m[tex]_{out[/tex]h[tex]_g[/tex] = m₂u[tex]_g[/tex] - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

In warm climates, a vapor barrier is placed on the exterior side of the insulation, and in cold climates it is installed on the interior side of the
insulation. Which of the following explains this placement of the barrier?
The barrier should always be placed on the side opposite from where the water condenses.
The barrier should always be placed on the side opposite where rain or snow hit.
The barrier should always be placed on the side where rain or snow hit.
The barrier should always be placed on the side where the water condenses

Answers

Answer: its c

Explanation:

In a parking garage, each level is 18 feet apart, Each ramp to a level is 120 feet long. Find the measurement of elevation for each ramp?

Answers

Answer:

Sin(angle)=0.1538

Angle=arcsine(0.1538)

Angle is 8.85 degrees

Explanation:

Firstly draw a diagram showing a triangle with the required angle, a hypotenuse of 130 ft and an opposite side of 20 ft.

The sine of the angle required uses the formula Opposite/Hypotenuse

(i am not sure sure)

what is the answer to life the universe and everything
(worth 95 points!)

Answers

i would say nothing, time is endless and the cycle of life is endless, not only on earth but almost anywhere, people try to find answers like what’s at the bottom of the ocean and stuff like that but there’s no point in finding out because it has no benefit, life is made for different reasons so there’s not one answer to it

Answer:

In the absence of dark energy, a flat universe expands forever but at a continually decelerating rate, with expansion asymptotically approaching zero; with dark energy, the expansion rate of the universe initially slows down, due to the effects of gravity, but eventually increases, and the ultimate fate of the universe ...

Explanation:

I think it goes on forever.

A cylindrical specimen of this alloy 12 mm in diameter and 188 mm long is to be pulled in tension. Assume a value of 0.34 for Poisson's ratio.Calculate the stress (in MPa) necessary to cause a 0.0105 mm reduction in diameter.

Answers

This question is incomplete, the missing image in uploaded along this answer below.

Answer:

The required stress is 200 Mpa

Explanation:

Given the data in the question;

diameter D = 12 mm = 12 × 10⁻³ m

Length L = 188 mm = 188 × 10⁻³ m

Poisson's ratio v = 0.34

Reduction in diameter Δd = 0.0105 mm = 0.0105 × 10⁻³ m

The transverse strain will;

εˣ = Δd / D

εˣ = -0.0105 × 10⁻³ /  12 × 10⁻³ m

εˣ = -0.00088

The longitudinal strain will be;

[tex]E^z[/tex] = - ( εˣ  / v )

[tex]E^z[/tex] = - ( -0.00088  / 0.34 )

[tex]E^z[/tex] = - ( - 0.002588 )

[tex]E^z[/tex] = 0.0026

Now, Using the values for strain, we get the value of stress from the graph provided in the question, ( first image uploaded below.

From the graph, in the Second image;

The stress is 200 Mpa

Therefore, The required stress is 200 Mpa

Which option identifies the step of the implementation phase represented in the following scenario?

A company in the gaming industry decides to integrate movement controls in its newest hardware release. It sends an online survey to interactive gamers to identify their highest priorities.

establishing a process and budget

using communication tools

building and assembling a team

setting up a change order process

Answers

Answer:

Which option identifies the step of the implementation phase represented in the following scenario?

A company in the gaming industry decides to integrate movement controls in its newest hardware release. It sends an online survey to interactive gamers to identify their highest priorities.

establishing a process and budget

using communication tools

building and assembling a team

setting up a change order process

Explanation:

#carryonml

Answer:

using communication tools

Explanation:

The correct answer is using communication tools. Communication tools such as online surveys help project teams identify customers’ wants and needs.

Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -16oC and a quality of 20% at a velocity of 5 m/s. At the exit, the refrigerant is a saturated vapor at -16oC. The evaporator flow channel has constant diameter of 1.7 cm. Determine the mass flow rate of the refrigerant, in kg/s, and the velocity at the exit, in m/s.

Answers

Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature [tex]T_1 = -16^0\ C[/tex]

Quality [tex]x_1 = 0.2[/tex]

Outlet:

Temperature [tex]T_2 = -16^0 C[/tex]

Quality  [tex]x_2 = 1[/tex]

The following data were obtained at saturation properties of R134a at the temperature of -16° C

[tex]v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\ v_g = 0.1247 \ m^3 /kg[/tex]

[tex]v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\ v_1 = 0.0255 \ m^3/kg \\ \\ \\ v_2 = v_g = 0.1247 \ m^3/kg[/tex]

[tex]m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6 \\ \\ \mathbf{m = 0.0534 \ kg/sec}[/tex]

[tex]\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2 \\ \\ \rho_1v_1 = \rho_2v_2 \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}[/tex]

A cylindrical metal specimen having an original diameter of 10.33 mm and gauge length of 52.8 mm is pulled in tension until fracture occurs. The diameter at the point of fracture is 6.38 mm, and the fractured gauge length is 73.9 mm. Calculate the ductility in terms of (a) percent reduction in area (percent RA), and (b) percent elongation (percent EL).

Answers

Answer

a) 62 percent

b) 40 percent

Explanation:

Original diameter ( d[tex]_{i}[/tex] ) = 10.33 mm

Original Gauge length ( L[tex]_{i}[/tex] ) = 52.8 mm

diameter at point of fracture ( d[tex]_{f}[/tex] ) = 6.38 mm

New gauge length ( L[tex]_{f}[/tex] ) = 73.9 mm

Calculate ductility in terms of

a) percent reduction in area

percentage reduction = [ (A[tex]_{i}[/tex] - A[tex]_{f}[/tex] ) / A[tex]_{i}[/tex] ] * 100

A[tex]_{i}[/tex] ( initial area ) = π /4 di^2

= π /4 * ( 10.33 )^2 = 83.81 mm^2

A[tex]_{f}[/tex] ( final area ) = π /4 df^2

= π /4 ( 6.38)^2 = 31.97 mm^2

hence : %reduction = ( 83.81 - 31.97 ) / 83.81

= 0.62 = 62 percent

b ) percent elongation

percentage elongation = ( L[tex]_{f}[/tex] - L[tex]_{i}[/tex] ) / L[tex]_{i}[/tex]

= ( 73.9 - 52.8 ) / 52.8 = 0.40 = 40 percent

In the context of mechanical systems, what does the term efficiency mean? OA the factor by which a machine multiplies a force B. the ratio of a machine's power to the force of its input Ос. the rate at which a machine performs work D. the rate at which a machine consumes energy E. the ratio of the work output of a machine to the work input​

Answers

Answer:

E

Explanation:

I have a big brain and I just took the test and got it correct.

An asphalt binder is mixed with aggregate and compacted into a sample. The mass of the dry sample is 1173.5 g, the mass of the sample submerged and then surface-dried with a damp towel is 1175.5 g, and the mass of the sample completely submerged in water is 652.5 g. Find the bulk specific gravity of the compacted sample.

Answers

Answer:

[tex]\mathbf{G_m = 2.25}[/tex]

Explanation:

From the given information:

Let the weight of the mix in the air be = [tex]W_{ma}[/tex]

Let the weight of the mix in water be = [tex]W_{mw}[/tex]; &

the bulk specific gravity be = [tex]G_m[/tex]

SO;

[tex]W_{mw} = W_{ma} - v \delta _{w} --- (1)[/tex]

Also;

[tex]G_m = \dfrac{W_{mw}}{v \delta_w} --- (2)[/tex]

From (2), make[tex]v \delta_w[/tex] the subject:

[tex]v \delta_w = \dfrac{W_{ma}}{G_m}[/tex]

Now, equation (1) can be rewritten as:

[tex]W_{mw} = W_{ma} - \dfrac{W_{ma}}{G_m}[/tex]

[tex]G_m = \dfrac{W_{ma}}{W_{ma} - W_{mw}}[/tex]

Replacing the values;

[tex]G_m = \dfrac{1173.5}{1173.5 -652.5}[/tex]

[tex]G_m = \dfrac{1173.5}{521}[/tex]

[tex]\mathbf{G_m = 2.25}[/tex]

A noisy transmission channel has a per-digit error probability p = 0.01.
(a) Calculate the probability of more than one error in 10 received digits?

Answers

Answer:

The appropriate answer is "0.0043".

Explanation:

The given values is:

Error probability,

p = 0.01

Received digits,

n = 10

and,

[tex]x\sim Binomial[/tex]

As we know,

⇒  [tex]P(x)=\binom{n}{x}p^xq^{n-x}[/tex]

Now,

⇒  [tex]P(x >1) =1- \left \{ P(x=0)+P(x=1) \right \}[/tex]

⇒                 [tex]=1-\left \{\binom{10}{0}(0.01)^0(0.99)^{10-0}+\binom{10}{0}(0.01)^1(0.99)^{10-1} \right \}[/tex]

⇒                 [tex]=1-0.9957[/tex]

⇒                 [tex]=0.0043[/tex]

MAPP gas, natural gas, propane, and acetylene can be used with oxygen to cut metal.
True
False

Answers

The answer is true.

Help meeeeeeeee plzzzzz need explanation

Answers

the picture is blank for me what does it say i can comment the answer plz mark brainlyist

As part of a heat treatment process, cylindrical, 304 stainless steel rods of 100-mm diameter are cooled from an initial temperature of 500 C by suspending them in an oil bath at 30 C. If a convection coefficient of 500 W/m2 K is maintained by circulation of the oil, how long does it take for the centerline of a rod to reach a temperature of 50 C, at which point it is withdrawn from the bath

Answers

Answer:

Explanation:

Given that:

diameter = 100 mm

initial temperature = 500 ° C

Conventional coefficient = 500 W/m^2 K

length  = 1 m

We obtain the following data from the tables A-1;

For the stainless steel of the rod [tex]\overline T = 548 \ K[/tex]

[tex]\rho = 7900 \ kg/m^3[/tex]

[tex]K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K[/tex]

[tex]\alpha = 4.40 \times 10^{-6} \ m^2/s \\ \\ B_i = \dfrac{h(\rho/4)}{K} \\ \\ =0.657[/tex]

Here, we can't apply the lumped capacitance method, since Bi > 0.1

[tex]\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\[/tex]

[tex]0.0426 = c_1 \ exp (- E^2_1 F_o_)\\ \\ \\ 0.0426 = 1.1382 \ exp (-10.9287)^2 \ f_o \\ \\ = f_o = \dfrac{In(0.0374)}{0.863} \\ \\ f_o = 3.81[/tex]

[tex]t_f = \dfrac{f_o r^2}{\alpha} \\ \\ t_f = \dfrac{3.81 \times (0.05)^2}{4.40 \times 10^{-6}} \\ \\ t_f= 2162.5 \\ \\ t_f = 36 mins[/tex]

However, on a single rod, the energy extracted is:

[tex]\theta = pcv (T_i - T_{\infty} )(1 - \dfrac{2 \theta}{c} J_1 (\zeta) ) \\ \\ = 7900 \\times 546 \times 0.007854 \times (500 -300) (1 - \dfrac{2 \times 0.0426}{1.3643}) \\ \\ \theta = 1.54 \times 10^7 \ J[/tex]

Hence, for centerline temperature at 50 °C;

The surface temperature is:

[tex]T(r_o,t) = T_{\infty} +(T_1 -T_{\infty}) \theta_o \ J_o(\zeta_1) \\ \\ = 30 + (500-30) \times 0.0426 \times 0.5386 \\ \\ \mathbf{T(r_o,t) = 41.69 ^0 \ C}[/tex]

You are performing a machining operation that approximates orthogonal cutting. Given that the chip thickness prior to chip formation is 0.5 inches and the chip thickness after separation is 1.125 inches, calculate the shear plane angle and shear strain. Use a rake angle of 10 degrees. 21. Suppose in the prior problem that the cutting force and thrust force were measured as 1559 N and 1271 N, respectively. The width of the cut is 3.0mm. Using this new information, calculate the shear strength of the material.

Answers

Answer:

A)

shear plane angle = 31.98°

shear strain = cot ( 31.98° ) + tan ( 31.98 - 10 )

B) shear strength = 7339.78

Explanation:

a) Determine the shear plane angle and shear strain

Given data :

Chip thickness before chip formation = 0.5 inches

Chip thickness after separation = 1.125 inches

rake angle ( ∝ ) = 10°  

shear plane angle : Tan ∅ = [tex]\frac{rcos\alpha }{1-sin\alpha }[/tex]    ----- ( 1 )

r = chip thickness ratio = 0.5 / 1.125 = 0.4444

back to equation 1 : Tan ∅ = ( 0.444 ) * cos 10 / 1 - sin 10

Tan ∅ = 0.4444 * 0.9848 / 1 - 0.1736  = 0.5296

hence ∅ = tan^-1 ( 0.5296 ) = 31.98°

shear strain :  R = cot ∅ + tan ( ∅ - ∝ ) ---------- ( 1 )

R = cot ( 31.98° ) + tan ( 31.98 - 10 )

  B) determine the shear strength of the material

cutting force = 1559 N

thrust force = 1271 N

width of cut ( diameter ) = 3.0 mm

shear strength = c + σ.tan ∅

c = cohesion force  = 1271 * 3  = 3813

σ = normal stress  = F / A = 1559 / π/4 * ( 0.5 )^2 = 1559 / 0.1963 = 7941.94

hence : shear strength of material = 3813 + 7941.94 * 0.6244 = 7339.78

A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elastic modulus of 103 GPa (15.0×106 psi). A cylindrical specimen of this alloy 5.4 mm (0.21 in.) in diameter and 225 mm (8.87 in.) long is stressed in tension and found to elongate 6.8 mm (0.27 in.). On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If not, explain why.

Answers

Answer:

The magnitude of the load can be computed  because it is mandatory in order to produce the change in length ( elongation )

Explanation:

Yield strength = 275 Mpa

Tensile strength = 380 Mpa

elastic modulus = 103 GPa

The magnitude of the load can be computed  because it is mandatory in order to produce the change in length ( elongation ) .

Given that the yield strength, elastic modulus and strain that is experienced by the test spectrum are given

strain = yield strength / elastic modulus

          = 0.0027

Technician A says when you push the horn button, electromagnetism moves an iron bar inside the horn, which opens and closes contacts in the horn circuit. Technician B says most vehicle horn circuits use a relay. Which technician is correct?

Answers

Answer: Both technicians A and B

Explanation:

I took the pf test

The evaporator:
A. directs airflow to the condenser.
B. absorbs heat from the passenger compartment.
C. removes moisture from the refrigerant.
D. restricts refrigerant flow.

Answers

Answer:

Option B

Explanation:

An evaporator along with  cold low pressure refrigerant absorbs heat from the air within the passenger compartment thereby supplying cool air for the occupants.

Hence, option B is correct

A tiger cub has a pattern of stripes on it for that is similar to that of his parents where are the instructions stored that provide information for a tigers for a pattern

Answers

probably in it's chromosomes

A 10 cm thick slab (density 8530 kg/m3 , specific heat 380J/kg K, conductivity 110 W/m K) initially at 650C is being cooled by air at 15C (convection coefficient 220 W/m2 K) at left surface. The right surface is insulated. We want to estimate temperature in the slab. a. What will be the temperature of the slab after a very long time

Answers

Answer:

The temperature after a long time will return to 15°C

Explanation:

Determine the temperature of the slab after a very long time

First we calculate the heat flow for m^2 area normal to the surface

= q / A = 650°c - 15°C / ( 1 / h  + L / K )

           = 635°c  / ( 1 / 220 + 0.1 / 110 )  = 116.416 kw/m^2

Total heat content in the slab is calculated as

= m* c * ΔT

= 8530 * A * 0.1 * 380 * ( 650 - 15 )

= 205828.9  kJ/m^2

The temperature will return to 15°C after a long time

Which of these parts reduces the amount of pollutants in the engine exhaust?
A. Transmission
B. Catalytic converter
C. Tailpipe
D. Muffler

Answers

Answer:

Option B

Explanation:

Catalytic convertor with in the exhaust pipes of any vehicle converts harmful gases (such as hydrocarbons (HC), carbon monoxide (CO) and nitrogen oxides (NOx)) produced by combustion of fossil fuels (petrol, diesel etc.) into less harmful/toxic pollutants by acting as a catalyst for redox reaction. A catalytic convertor releases some mild gases from the exhaust like less harmful CO2, nitrogen and water vapour (steam)

Hence, option B is correct

5. The pin support at A allows _______. Select the one that applies. (a) displacement in the x direction (b) rotation about its central axis (c) displacement in the y direction (d) none of the above 6. The support at B does not allow _______. Select the one that applies. (a) displacement in the x direction (b) rotation about its central axis (c) displacement in the y direction

Answers

Answer: Diagram associated with your question is attached below

5) B

6) C

Explanation:

5) The pin support at A allows ; Rotation about its central axis  

This is because pin supports does not allow the translation of its structural member in any direction i.e. y or x but only rotation about its axis

6)  The support at B does not allow displacement in y direction

This is because roller support allows displacement only in the direction that they are situated and in this case it is the x - direction

Some project managers prefer the PERT chart over the Gantt chart because it clearly illustrates task dependencies. A PERT chart, however, can be much more difficult to interpret, especially on complex projects. Alternatively, some project managers may choose to use both techniques. If you are the project manager of a residential construction project, will you prefer PERT chart to Gantt chart? Explain why?

Answers

Answer:

PERT Chart and GANTT Chart

As the project manager of a residential construction project, I will prefer the PERT chart to the GANTT chart because a PERT chart displays task dependencies unlike a Gantt chart.  With the PERT chart, the sequence of tasks is clearly mapped out.  Dependent tasks are carried out when other tasks that they depend on have been executed.

Explanation:

By definition, a Gantt chart is like a bar chart that lays out project tasks and timelines using bars.  On the other hand, a PERT chart follows a structure in the form of flow charts or network diagrams.  It displays all the project tasks in separate boxes.  The boxes are then connected with arrows which clearly show the task dependencies.

In the production of soybean oil, dried and flaked soybeans are brought in contact with a solvent (often hexane) that extracts the oil and leaves behind the residual solids and a small amount of oil.

a. Draw flow diagram of the process, labeling the two feed streams (beans and solvent) and the leaving streams (solids and extract).
b. The soybeans contain 18.5 wt% oil and the remainder insoluble solids, and the hexane is fed at a rate corresponding to 2.0 kg hexane per kg beans. The residual solids leaving the extraction unit contain 35.0 wt% hexane, all of the non-oil solids that entered with beans, and 1.0% of the oil that entered the beans. For a feed rate of 1000 kg/h of dried flaked soybeans, calculate mass flow rates of extract and residual solids and the composition of extract.

Answers

Answer: its c

Explanation:

Design a counter that counts the following sequence of 2-0-1-3 and repeat.

Answers

Answer:

Hello your question is incomplete below is the complete question

Design a counter that counts the following sequence of 2-0-1-3 and repeat. Use the JK flip-flops given to you at the start of the semester. These values will be displayed on a seven-segment display like the one used in Lab 3

answer : attached below

Explanation:

Designing a counter that counts in a given sequence can be done using Logic gates that will be used to control the counter. we will design the counter using the 7 segment display

Note : The first Image is the segment display and the second image is the design of the counter

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