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Juli2301 [7.4K]

11 months ago

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A pipe of length 10.0 m increases in length by 1.5 cm when its temperature is increased by 90°F. What is its coefficient of line

ar expansion?

1 answer:

azamat11 months ago

3 0

The coefficient of linear expansion, given that the length of the pipe increased by 1.5 cm is 1.67×10¯⁵ /°F

<h3>How to determine the coefficient of linear expansion</h3>

From the question given above, the following data were obtained

Original diameter (L₁) = 10 m
Change in length (∆L) = 1.5 cm = 1.5 / 100 = 0.015 m
Change in temperature (∆T) = 90 °F
Coefficient of linear expansion (α) =?

The coefficient of linear expansion can be obtained as illustrated below:

α = ∆L / L₁∆T

α = 0.015 / (10 × 90)

α = 0.015 / 900

α = 1.67×10¯⁵ /°F

Thus, we can conclude that the coefficient of linear expansion is 1.67×10¯⁵ /°F

Learn more about coefficient of linear expansion:

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Moist air initially at 1258C, 4 bar, and 50% relative humidity is contained in a 2.5-m3 closed, rigid tank. The tank contents ar

brilliants [131]

Here is the missing part of the question

To Determine the heat transfer, in kJ if the final temperature in the tank is 110 deg C

Answer:

Explanation:

The image attached below shows the process on T - v diagram

<u>At State 1:</u>

The first step is to find the vapor pressure

$P_{v1} = \rho_1 P_g_1$

$= \phi_1 P_{x \ at \ 125^0C}$

= 0.5 × 232 kPa

= 116 kPa

The initial specific volume of the vapor is:

$P_{v_1} v_{v_1} = \dfrac{\overline R}{M_v}T_1$

$116 \times 10^3 \times v_{v_1} = \dfrac{8314}{18} \times (125 + 273)$

$116 \times 10^3 \times v_{v_1} = 183831.7778$

$v_{v_1} = 1.584 \ m^3/kg$

<u>At State 1:</u>

The next step is to determine the mass of water vapor pressure.

$m_{v1} = \dfrac{V}{v_{v1}}$

$= \dfrac{2.5}{1.584}$

= 1.578 kg

Using the ideal gas equation to estimate the mass of the dry air $m_a$ $P_{a1} V = m_a \dfrac{\overline R}{M_a}T_1$

$(P_1-P_{v1}) V = m_a \dfrac{\overline R}{M_a}T_1$

$(4-1.16) \times 10^5 \times 2.5 = m_a \dfrac{8314}{28.97}\times ( 125 + 273)$

$710000= m_a \times 114220.642$

$m_a = \dfrac{710000}{114220.642}$

$m_a = 6.216 \ kg$

For the specific volume $v_{v_1} = 1.584 \ m^3/kg$ , we get the identical value of saturation temperature

$T_{sat} = 100 + (110 -100) \bigg(\dfrac{1.584-1.673}{1.210 - 1.673}\bigg)$

$T_{sat} =101.92 ^0\ C$

Thus, at $T_{sat} =101.92 ^0\ C$ , condensation needs to begin.

However, since the exit temperature tends to be higher than the saturation temperature, then there will be an absence of condensation during the process.

Heat can now be determined by using the formula

Q = ΔU + W

Recall that: For a rigid tank, W = 0

Q = ΔU + 0

Q = ΔU

Q = U₂ - U₁

Also, the mass will remain constant given that there will not be any condensation during the process from state 1 and state 2.

<u>At State 1;</u>

The internal energy is calculated as:

$U_1 = (m_a u_a \ _{ at \ 125^0 C})+ ( m_{v1} u_v \ _{ at \ 125^0 C} )$

At $T_1$ = 125° C, we obtain the specific internal energy of air

SO;

$U_{a \ at \ 125 ^0C } = 278.93 + ( 286.16 -278.93) (\dfrac{398-390}{400-390} )$

$=278.93 + ( 7.23) (\dfrac{8}{10} )$

$= 284.714 \ kJ/kg\\$

At $T_1$ = 125° C, we obtain the specific internal energy of water vapor

$U_{v1 \ at \ 125^0C} = u_g = 2534.5 \ kJ/kg$

$U_1 = (m_a u_a \ at \ _{ 125 ^0C }) + ( m_{v1} u_v \ at \ _{125^0C} )$

= 6.216 × 284.714 + 1.578 × 2534.5

= 5768.716 kJ

<u>At State 2:</u>

The internal energy is calculated as:

$U_2 = (m_a u_a \ _{ at \ 110^0 C})+ ( m_{v1} u_v \ _{ at \ 110^0 C} )$

At temperature 110° C, we obtain the specific internal energy of air

SO;

$U_{a \ at \ 110^0C } = 271.69+ ( 278.93-271.69) (\dfrac{383-380}{390-380} )$

$271.69+ (7.24) (0.3)$

$= 273.862 \ kJ/kg\\$

At temperature 110° C, we obtain the specific internal energy of water vapor

$U_{v1 \ at \ 110^0C}= 2517.9 \ kJ/kg$

$U_2 = (m_a u_a \ at \ _{ 110 ^0C }) + ( m_{v1} u_v \ at \ _{110^0C} )$

= 6.216 × 273.862 + 1.578 × 2517.9

= 5675.57 kJ

Finally, the heat transfer during the process is

Q = U₂ - U₁

Q = (5675.57 - 5768.716 ) kJ

Q = -93.146 kJ

with the negative sign, this indicates that heat is lost from the system.

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2 years ago

Briefly describe the three intelligences included in Robert Sternberg’s triarchic theory of human intelligence.

pshichka [43]

The triarchic theory of intelligence<span> was formulated by </span>Robert J. Sternberg<span>, a prominent figure in research of human </span>intelligence<span>. The theory by itself was among the first to go against the </span>psychometric<span> approach to intelligence and take a more </span>cognitive approach<span>. The three meta components are also called triarchic components. These are the triarchic theory of human intelligence.
</span>1. Analytical - Analytical Intelligence similar to the standard psychometric definition of intelligence e.g. as measured by Academic problem solving: analogies and puzzles, and corresponds to his earlier componential intelligence. Sternberg considers this reflects how an individual relates to his internal world.

Sternberg believes that Analytical Intelligence (Academic problem-solving skills) is based on the joint operations of metacomponents and performance components and knowledge acquisition components of intelligence

2. Practical - Practical Intelligence: this involves the ability to grasp, understand and deal with everyday tasks. This is the Contextual aspect of intelligence and reflects how the individual relates to the external world about him or her.

<span>Sternberg states that Intelligence is: </span>"Purposive adaptation to, shaping of, and selection of real-world environments relevant to one's life" (Sternberg, 1984, p.271)

3. Creative - Creative Intelligence: this involves insights, synthesis and the ability to react to novel situations and stimuli. This he considers the Experiential aspect of intelligence and reflects how an individual connects the internal world to external reality.

<span>Sternberg </span>considers the Creative facet to consist of the ability which allows people to think creatively and that which allows people to adjust creatively and effectively to new situations.

<span>Sternberg believes that more intelligent individuals will also move from consciously learning in a novel situation to automating the new learning so that they can attend to other tasks.</span>

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3 years ago

C. A helium atom has a diameter of approximately 9.8 • 10-11 meters. What is the diameter of a helium atom in nanometers? Given

Studentka2010 [4]

$1\ nm = 10^{-9} m\\\\1\ m =\frac{1\ nm}{10^{-9}}$

Since the diameter of helium atom is approximately $9.8\times 10^{-11}\ m$ , therefore the diameter of helium atom in nano meter,

$9.8\times 10^{-11}\ m=9.8\times 10^{-11} \times\frac{1\ nm}{10^{-9}}=0.098\ nm$

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3 years ago

"A block of metal weighs 40 N in air and 30 N in water. What is the buoyant force on the block due to the water? The density of

Alja [10]

Answer:

buoyant force on the block due to the water= 10 N

Explanation:

We know that

buoyant force(F_B) on a block= weight of the block in air (actual weight) - weight of block in water.

Given:

A block of metal weighs 40 N in air and 30 N in water.

F_B = 40-30= 10 N

therefore, buoyant force on the block due to the water= 10 N

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2 years ago

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A person has a mass of 60 kg. What is the person’s weight in Newtons and in pounds?

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Answer:

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