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mash [69]
3 years ago
9

A block of aluminum occupies a volume of 15.0 ml and weighs 40.5 what is its density

Physics
2 answers:
satela [25.4K]3 years ago
6 0
40.5 / 15 =
2.7 g/ml
AVprozaik [17]3 years ago
5 0
D = 40.5 g / 15.0 mL<span>d = 2.70 g/mL</span>
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A spring has a spring constant of 90N/m.How much potential energy does it store when stretched by 2 cm?
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Answer:

The potential energy stored in the spring is 0.018 J.

Explanation:

Given;

spring constant, k = 90 N/m

extension of the spring, x = 2 cm = 0.02 m

The potential energy stored in the spring is calculated as;

U = ¹/₂kx²

where;

U is the potential energy stored in the spring

Substitute the given values in the equation above;

U = ¹/₂  x  90 N/m  x  (0.02 m)²

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Water enters the turbine of an ideal Rankine cycle as a superheated vapor at 18 MPa and 550°C. If the condenser pressure is 9 kP
ivolga24 [154]

Answer:

For this ideal Rankine cycle:

A) The rate of heat addition into the cycle Qh is 3214.59KJ/Kg.

B) The thermal efficiency of the system εt=0.4354.

C) The rate of heat rejection from the cycle Qc is 1833.32KJ/Kg.

D) The back work ratio is 0.013

Explanation:

To solve the ideal Rankine cycle we have to determinate the thermodynamic information of each point of the cycle. We will use a water thermodynamic properties table. In an ideal Rankine cycle, the process in the turbine and the pump must be isentropic. Therefore S₃=S₄ and S₁=S₂.

We will start with the point 3:

P₃=18000KPa  T₃=550ºc ⇒ h₃=3416.12 KJ/Kg  S₃=6.40690 KJ/Kg

Point 4:

P₄=9KPa S₄=S₃ ⇒ h₄=2016.60 (x₄=0.7645 is wet steam)

Point 1:

P₁=P₄ in the endpoint of the steam curve. ⇒ h₁= 193.28KJ/Kg  S₁=0.62235KJ/Kg  x₁=0

Point 2:

P₂=P₃ and S₂=S₁ ⇒ h₂=201.53KJ/Kg

With this information we can obtain the heat rates, the turbine, and the pump work:

Q_h=Q_{2-3}=h_3-h_2=3214.59KJ/Kg

Q_c=Q_{4-1}=h_4-h_1=1833.32KJ/Kg

W_{turb}=W_{3-4}=h_3-h_4=1399.52KJ/Kg

W_{pump}=W_{1-2}=h_2-h_1=18.25KJ/Kg

We can answer the questions with this data:

A) Q_h=Q_{2-3}=h_3-h_2=3214.59KJ/Kg

B) \epsilon_{ther}=\displaystyle\frac{W_{turb}}{Q_h}=0.4354

C)Q_c=Q_{4-1}=h_4-h_1=1833.32KJ/Kg

D) r_{bW}=\displaystyle\frac{W_{pump}}{W_{turb}}=0.013

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B. Pairs

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