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Naddika [18.5K]
3 years ago
14

For the system form of the basic laws, the momentum of a system can change as a result of: a. pressure acting on the system. b.

friction acting on the system. c. gravity acting on the system. d. body forces acting on the system. e. surface forces acting on the system. f. all of the above.
Engineering
2 answers:
Ymorist [56]3 years ago
7 0

Answer: F. All the above.

Explanation:

Basically, if an object is moving, it moves with a certain velocity and mass. Momentum of a body is a product of mass and velocity. The sum of momentum of individual bodies is equal to the entire system momentum. For instance, a block v is moving due to an applied force F, with a velocity V and the gravity g. Due to gravity, the weight is mg. Due to gravity, the weight is acting downward. Applied force is acting on the block surface area A.

Pressure = Surface force/Surface area.

Surface force is acting on the surface applied. So, acting pressure = Force applied/area.

Pressure and surface force is acting on the body. Body forces also act on the system. Forces due to gravity is also referred to as body force. As a result of weight of the box, Normal force produced by the rough surface is equal to the Weight. As a result of rough surface, frictional forces are produced which opposes the block to move forward. All the external forces create a net total force due to which the block move with a velocity and acceleration.

In Newton's second law, Ftotal is equal to mass × acceleration.

Therefore, we can conclude that momentum can change as a result of all these forces because mass × acceleration is related to total force and momentum is equal to mass × velocity.

sammy [17]3 years ago
7 0

Answer:

Option f

All of the above

Explanation:

Considering a control volume, the system form shows that the change in momentum is affected by any force at all on the system, this includes, pressure force, frictional force, gravity force, or any other body or surface forces on the system.

For steady flow with a fixed control volume, the most useful form of the momentum equation is given as:

\Sigma F_{gravity}+\Sigma F_{pressure}+\Sigma F_{viscosity}+\Sigma F_{other}=\Sigma mV_{av}

This shows that all forces, whether body forces or contact forces present in the system can affect the momentum of the system.

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The inlet and exhaust flow processes are not included in the analysis of the Otto cycle. How do these processes affect the Otto
lara31 [8.8K]

Answer:

Suction and exhaust processes do not affect the performance of Otto cycle.

Explanation:

Step1

Inlet and exhaust flow processes are not including in the Otto cycle because the effect and nature of both the process are same in opposite direction.

Step2

Inlet process or the suction process is the process of suction of working fluid inside the cylinder. The suction process is the constant pressure process. The exhaust process is the process of exhaust out at constant pressure.

Step3

The suction and exhaust process have same work and heat in opposite direction. So, net effect of suction and exhaust processes cancels out. The suction and exhaust processes are shown below in P-V diagram of Otto cycle:

Process 0-1 is suction process and process 1-0 is exhaust process.

7 0
3 years ago
Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans
aksik [14]

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, T_{i} = 20^{\circ}C = 273 + 20 = 293 K

Temperature at the outlet, T_{o} = 200{\circ}C = 273 + 200 = 473 K

Pressure at inlet, P_{i} = 80 kPa = 80\times 10^{3} Pa

Pressure at outlet, P_{o} = 800 kPa = 800\times 10^{3} Pa

Speed at the outlet, v_{o} = 20 m/s

Diameter of the tube, D = 10 cm = 10\times 10^{- 2} m = 0.1 m

Input power, P_{i} = 400 kW = 400\times 10^{3} W

Now,

To calculate the heat transfer, Q, we make use of the steady flow eqn:

h_{i} + \frac{v_{i}^{2}}{2} + gH  + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}

where

h_{i} = specific enthalpy at inlet

h_{o} = specific enthalpy at outlet

v_{i} = air speed at inlet

p_{s} = specific power input

H and H' = Elevation of inlet and outlet

Now, if

v_{i} = 0 and H = H'

Then the above eqn reduces to:

h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}

Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}                (1)

Also,

p_{s} = \frac{P_{i}}{ mass, m}

Area of cross-section, A = \frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}

Specific Volume at outlet, V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s

From the eqn:

P_{o}V_{o} = mRT_{o}

m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s

Now,

p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg

Also,

\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg

Now, using these values in eqn (1):

Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW

Now, rate of heat transfer, q:

q = mQ = 0.925\times 813.33 = 752.33 kW

4 0
3 years ago
Air enters a compressor operating at steady state at 1 bar, 290 K, with a mass flow rate of 0.1 kg/s and exits at 980 K, 10 bar.
stiv31 [10]

Answer:

7.615 kW

Explanation:

Solution in pen paper form in the attachment section

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3 years ago
Why is using the proper joining technique important? What could go wrong if the wrong joining technique is used?
Alex73 [517]

Answer:

Explanation:

Using the proper technique is incredibly important because it prevents the materials being joined from breaking and/or causing an accident. If the wrong joining technique is used the materials may not hold in place and come apart easily instead. Also, some joining techniques are not meant for some materials and may instead cause the material to become weak and brittle causing it to break apart almost immediately.

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The yield stress of a steel is 250Mpa. A steel rod used for implant in a femurneeds to withstand 29KN. What should the diameter
OleMash [197]

Answer:

r = 1.922 mm

Explanation:

We are given;

Yield stress; σ = 250 MPa = 250 N/mm²

Force; F = 29 KN = 29000 N

Now, formula for yield stress is;

σ = F/A

A = F/σ

Where A is area = πr²

Thus;

r² = 2900/250π

r² = 3.6924

r = √3.6924

r = 1.922 mm

3 0
3 years ago
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