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Alenkinab [10]
3 years ago
15

Define Mechanism and mechanics.​

Engineering
2 answers:
aivan3 [116]3 years ago
6 0

mechanism, in mechanical construction, the means employed to transmit and modify motion in a machine or any assemblage of mechanical parts.

VashaNatasha [74]3 years ago
5 0

Answer:

MECHANISM-mechanism, in mechanical construction, the means employed to transmit and modify motion in a machine or any assemblage of mechanical parts. The chief characteristic of the mechanism of a machine is that all members have constrained motion; i.e., the parts can move only in a determinate manner relative to one another.

MECHANICS-The definition of mechanics is the branch of physics that deals with how the action of force affects material bodies or the functional parts of an activity.

An example of mechanics is the distance it takes for a car going 30mph to stop.

Explanation:

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What is the IMA of a fixed pulley ​
Amanda [17]

Answer:

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Explanation:

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3 years ago
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Water is contained in a rigid vessel of 5 m3 at a quality of 0.8 and a pressure of 1 MPa. If the pressure is reduced to 270.3 kP
professor190 [17]

Answer:

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Explanation:

4 0
3 years ago
Two mass streams of the same ideal gas are mixed in a steady-flow chamber while receiving energy by heat transfer from the surro
loris [4]

Answer:

(a)The final temperature of mixture is T₃ =m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp

(b) The final volume is V₃ =V₁ + V₂ + RQin/P₃Cp

(c) The volume flow rate at exit is V₃ =V₁ + V₂

Explanation:

Solution

Now

The system comprises of two inlets and on exit.

Mass flow rate enthalpy of fluid from inlet -1 be m₁ and h₁

Mass flow rate enthalpy of fluid from inlet -2 be m₂ and h₂

Mass flow rate enthalpy of fluid from  exit be m₃ and h₃

Mixing chambers do not include any kind of work (w = 0)

So, both  the kinetic and potential energies of the fluid streams are usually negligible (ke =0, pe =0)

(a) Applying the mass balance of mixing chamber, min = mout

Applying the energy balance of mixing chamber,

Ein = Eout

min hin =mout hout

miCpT₁ + m₂CpT₂ +Qin =m₃CpT₃

T₃ = miCpT₁/m₃CpT₃ + m₂CpT₂/m₃CpT₃ + Qin/m₃CpT₃ +

T₃ =m₁T₁/m₃+ m₂T/m₃ + Qin/m₃Cp

The final temperature of mixture is T₃ =m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp

(b) From the ideal gas equation,

v =RT/PT

v₃ = RT₃/P₃

The volume flow rate at the exit, V₃ =m₃v₃

V₃ = m₃ RT₃/P₃

Substituting the value of T₃, we have

V₃=m₃ R/P₃ (=m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp)

V₃ =  R/P₃ (m₁T₁+ m₂T₂ + Qin/Cp)

Now

The mixing process occurs at constant pressure P₃=P₂=P₁.

Hence V₃ becomes:

V₃=m₁RT₁/P₁ +m₂RT₂/P₂ + RQin/P₃Cp

V₃ =V₁ + V₂ + RQin/P₃Cp

Therefore, the final volume is V₃ =V₁ + V₂ + RQin/P₃Cp

(c) Now for an adiabatic mixing, Qin =0

Hence V₃ becomes:

V₃ =V₁ + V₂ + r * 0/P₃Cp

V₃ =V₁ + V₂ + 0

V₃ =V₁ + V₂

Therefore the volume flow rate at exit is V₃ =V₁ + V₂

8 0
3 years ago
Three single-phase, 10 kVA, 2400/280 V, 60-Hz transformers are connected to form a three-phase, 2400/480 V transformer The equiv
Dominik [7]
The answer to this question is letter A
7 0
3 years ago
Determine the angular acceleration of the uniform disk if (a) the rotational inertia of the disk is ignored and (b) the inertia
lukranit [14]

Answer:

α = 7.848 rad/s^2  ... Without disk inertia

α = 6.278 rad/s^2  .... With disk inertia

Explanation:

Given:-

- The mass of the disk, M = 5 kg

- The right hanging mass, mb = 4 kg

- The left hanging mass, ma = 6 kg

- The radius of the disk, r = 0.25 m

Find:-

Determine the angular acceleration of the uniform disk without and with considering the inertia of disk

Solution:-

- Assuming the inertia of the disk is negligible. The two masses ( A & B )  are hung over the disk in a pulley system. The disk is supported by a fixed support with hinge at the center of the disk.

- We will make a Free body diagram for each end of the rope/string ties to the masses A and B.

- The tension in the left and right string is considered to be ( T ).

- Apply newton's second law of motion for mass A and mass B.

                      ma*g - T = ma*a

                      T - mb*g = mb*a

Where,

* The tangential linear acceleration ( a ) with which the system of two masses assumed to be particles move with combined constant acceleration.

- g: The gravitational acceleration constant = 9.81 m/s^2

- Sum the two equations for both masses A and B:

                      g* ( ma - mb ) = ( ma + mb )*a

                      a =  g* ( ma - mb ) / ( ma + mb )

                      a = 9.81* ( 6 - 4 ) / ( 6 + 4 ) = 9.81 * ( 2 / 10 )

                      a = 1.962 m/s^2  

- The rope/string moves with linear acceleration of ( a ) which rotates the disk counter-clockwise in the direction of massive object A.

- The linear acceleration always acts tangent to the disk at a distance radius ( r ).

- For no slip conditions, the linear acceleration can be equated to tangential acceleration ( at ). The correlation between linear-rotational kinematics is given below :

                     a = at = 1.962 m/s^2

                     at = r*α      

Where,

           α: The angular acceleration of the object ( disk )

                    α = at / r

                    α = 1.962 / 0.25

                    α = 7.848 rad/s^2                                

- Take moments about the pivot O of the disk. Apply rotational dynamics conditions:

             

                Sum of moments ∑M = Iα

                 ( Ta - Tb )*r = Iα

- The moment about the pivots are due to masses A and B.

 

               Ta: The force in string due to mass A

               Tb: The force in string due to mass B

                I: The moment of inertia of disk = 0.5*M*r^2

                   ( ma*a - mb*a )*r = 0.5*M*r^2*α

                   α = ( ma*a - mb*a ) / ( 0.5*M*r )

                   α = ( 6*1.962 - 4*1.962 ) / ( 0.5*5*0.25 )

                   α = ( 3.924 ) / ( 0.625 )

                   α = 6.278 rad/s^2

6 0
3 years ago
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