2. On a trip from Alphaville to Betaville, you can take two main routes. Route 1, which goes through Gammatown, is 50 kilometers [km] long; however, you can only drive an average speed of 36 miles per hour [mph]. Route 2 travels along the freeway, at an average speed of 50 mph, but it is 65 km long. How long does it take to complete each route? State the time for each route in minutes [min].

Answer:

C. nylon

Explanation:

because it is definitely stronger than PLA And they are all not the same and acrylic isn’t strong than nylon

It should benotedtha for 100, repeat distance =ao=0.35167 nm

linear density=1/ao=2.84 points/nm

linear packing fraction =(2)(0.1243)(2.84) =0.707

For 110

repeat distance = 2*ao/2 = 0.2487 nm

linear density = 2 /( sqrt{2} * ao ) = 4.021 points/nm

linear packing fraction = (2)(0.1243)(4.02) =1.0

For 111,

repeat distance =(sqrt{3} * ao) = 0.6091 nm

linear density = 1 / (sqrt{3} * ao ) = 1.642 points/nm

linear packing fraction = (2)(0.1243)(1.642) =0.408

"Repeat distance" refers to the distance between repeated units in a crystal lattice. This can be measured along any of the lattice vectors that define the periodicity of the crystal. For example, in a simple cubic crystal, the repeat distance along any of the three lattice vectors would be equal to the length of that vector.

"Linear density" refers to the number of atoms or ions per unit length along a specific crystallographic direction. It is usually expressed in units of atoms/angstrom or ions/angstrom. To calculate the linear density, one would count the number of atoms or ions along a given direction and divide by the length of that direction.

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Answer:

Health informatics careers combine concepts and practices from healthcare, computer science, and information science. Professionals in this field can work as computer systems managers, health informationists, and clinical analysts. They can also pursue careers in higher education and applied research.

a) Note that the final volume of the air is also 1.6 m3, and the final pressure is 4.5 bar; and

b)  the change in internal energy of the air is 0 kJ, the work done by the air is 0.966 kJ, and the total heat transferred is 487.7 kJ.

To solve this problem, we can use the ideal gas law and the first law of thermodynamics.

(a) To determine the final volume and final pressure of the air, we can use the ideal gas law, which states that:

P1V1/T1 = P2V2/T2

where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.

We can first use this equation to determine the final volume of the air during the isothermal process:

P1V1/T1 = P2V2/T2

(9 bar)(0.8 m3)/(1200 K) = P2V2/(927 K)

V2 = (9/4.5) * (0.8 m3)
= 1.6 m3

Now, we can use the ideal gas law again to determine the final pressure of the air during the isothermal process:

P1V1/T1 = P2V2/T2

(9 bar)(0.8 m3)/(1200 K) = (P2)(1.6 m3)/(1200 K)

P2 = (9 bar)(0.8 m3)/(1.6 m3)
= 4.5 bar

During the subsequent heat transfer process, the piston is fixed in place, which means the volume of the air remains constant at 1.6 m3.

Therefore, the final volume of the air is also 1.6 m3, and the final pressure is 4.5 bar.

(b) To determine the change in internal energy of the air, the work done by the air, and the total heat transferred, we can use the first law of thermodynamics, which states that:

ΔU = Q - W

where ΔU is the change in internal energy, Q is the heat transferred, and W is the work done.

During the isothermal process, the temperature of the air remains constant at 927 K, which means the internal energy of the air does not change:

ΔU = 0

During the isothermal process, the work done by the air is:

W = -P1V1 ln(P2/P1)
= -(9 bar)(0.8 m3) ln(4.5/9)

= 0.966 kJ

During the subsequent heat transfer process, the air is heated from 927 K to 300 K, which means heat is transferred into the air:

Q = m cΔT = (P2V2/R) cΔT
= (4.5 bar)(1.6 m3)/(287 J/kg∙K) (1.005 kJ/kg∙K) (927 K - 300 K)
= 487.7 kJ

Therefore, the change in internal energy of the air is 0 kJ, the work done by the air is 0.966 kJ, and the total heat transferred is 487.7 kJ.

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The tension in the rope and the spring balance reading are the same. 5N of tension is present on both sides. As a result, the spring balance reads 5N.

A simple and affordable way to determine mass is by using spring balances. The mass is suspended from the end of a spring, and the spring's deflection as a result of the mass's downward gravitational pull is measured in relation to a scale.

A spring balance calculates an object's weight by comparing the force of gravity to the force of an extended spring. The scale can be used to calculate weight and mass by recalibrating it. Some spring balances have gramme or kilogram marks and are used to determine an object's mass.

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