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erica [24]
4 years ago
5

State whether the following statements are true or false. Provide a brief justification for your answer.

Engineering
1 answer:
Lena [83]4 years ago
5 0

Answer:

[a] False.

[b]. True

[c]. false.

[d]. true.

[e]. true.

Explanation:

NB: SDOF simply means Simple Degree Of Freedom, that is to say it is a system that can be solved by differential equation such as the second order and the single differential equation.

So, the question asked us to determine if each of the scenario is true or false.  

a. Consider a SDOF system with Coulomb damping. When displaced from the equilibrium position and released, the mass may not move at all.

<u><em>ANSWER:</em></u>  FALSE.

REASON: The mass moved a little bit When displaced from the equilibrium position and released.

b. Consider a SDOF system with an ideal viscous damper. When displaced from the equilibrium position and released, the mass will always undergo oscillatory motion.

<em><u>ANSWER:</u></em> TRUE

<em><u>REASON:  for an ideal viscous damper. When displaced from the equilibrium position and released, the mass will always undergo oscillatory motion.</u></em>

c. The damped natural frequency of a system is always greater than the undamped natural frequency

.

<em><u>ANSWER:  </u></em>FALSE

<em><u>REASON:  The damped natural frequency of a system is always greater than the undamped natural frequency</u></em>

d. For an undamped system undergoing a harmonic forcing, the amplitude of the response approaches zero as the forcing frequency becomes very high.

<em><u>ANSWER:</u></em> TRUE

<em><u>REASON:  the amplitude of the response approaches zero as the forcing frequency becomes very high for  undamped system undergoing a harmonic forcing</u></em>

e. The amplitude of free response of SDOF system with Coulomb damping decreases

<em><u>ANSWER:</u></em> TRUE

<em><u>REASON:  amplitude of free response of SDOF system with Coulomb damping decreases</u></em>

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Rhe fact that Didymos B is in orbit around Didymos A makes it easier to see the results of the impact, and ensures that the experiment doesn't change the orbit of the pair around the sun.” After launch, DART would fly to Didymos, and use an on-board autonomous targeting system to aim itself at Didymos B

Explanation:

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Technician A says that a circuit with continuity reads 0 ohms. Technician B says that an open circuit reads 0 ohms. Who is corre
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Answer:

  A

Explanation:

An open circuit has infinite resistance, not 0. A circuit with continuity usually measures a few ohms or less, depending on the length and gauge of the wire and the condition of the connections. Often, on a general-purpose ohmmeter, the reading is very near 0 ohms.

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8 0
2 years ago
A refrigerated space is maintained at -15℃, and cooling water is available at 30℃, the refrigerant is ammonia. The refrigeration
Illusion [34]

Answer:

(1) 5.74

(2) 5.09

(3) 3.05×10⁻⁵ kg/s

(4) 0.00573 kW

Explanation:

The parameters given are;

Working temperature, T_C  = -15°C = 258.15 K

Temperature of the cooling water, T_H = 30°C = 303.15 K

(1) The Carnot coefficient of performance is given as follows;

\gamma_{Max} = \dfrac{T_C}{T_H - T_C}  =  \dfrac{258.15}{303.15 - 258.15}   = 5.74

(2) For ammonia refrigerant, we have;

h_2 = h_g = 1466.3 \ kJ/kg

h_3 = h_f = 322.42 \ kJ/kg

h_4 = h_3 = h_f = 322.42 \ kJ/kg

s₂ = s₁ = 4.9738 kJ/(kg·K)

0.4538 + x₁ × (5.5397 - 0.4538) = 4.9738

∴ x₁ = (4.9738 - 0.4538)/(5.5397 - 0.4538) = 0.89

h_1 = h_{f1} + x_1 \times h_{gf}

h₁ = 111.66 + 0.89 × (1424.6 - 111.66) = 1278.5 kJ/kg

\gamma = \dfrac{h_1 - h_4}{h_2 - h_1}

\gamma = \dfrac{1278.5 - 322.42}{1466.3 - 1278.5} = 5.09

(3) The circulation rate is given by the mass flow rate, \dot m as follows

\dot m = \dfrac{Refrigeration \ capacity}{Refrigeration \ effect \ per \ unit \ mass}

The refrigeration capacity = 105 kJ/h

The refrigeration effect, Q = (h₁ - h₄) = (1278.5 - 322.42) = 956.08 kJ/kg

Therefore;

\dot m = \dfrac{105}{956.08}  = 0.1098 \ kg/h

\dot m = 0.1098 kg/h = 0.1098/(60*60) = 3.05×10⁻⁵ kg/s

(4) The work done, W = (h₂ - h₁) = (1466.3 - 1278.5) = 187.8 kJ/kg

The rating power = Work done per second = W×\dot m

∴ The rating power = 187.8 × 3.05×10⁻⁵ = 0.00573 kW.

6 0
3 years ago
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