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Temka [501]
3 years ago
5

What is described by the terms body-centered cubic and face-centered cubic?

Physics
1 answer:
Montano1993 [528]3 years ago
7 0
In physical chemistry, the terms body-centered cubic  (BCC) and face-centered cubic (FCC) refer to the cubic crystal system of a solid. Each solid is made up simple building blocks called lattice units. There are different layouts of a lattice unit.

It is better understood using 3-D models shown in the picture. A BCC unit cell has one lattice point in the center, together with eight corner atoms which represents 1/8 of an atom. Therefore, there are 1+ 8(1/8) = 2 atoms in a BCC unit cell. On the other hand, a FCC unit cell is composed of half of an atom in each of its faces and 1/8 of an atom in its corners. Therefore, there are (1/2)6 + (1/8)8 = 4 atoms in a FCC unit cell.

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Describe the mechanical energy of a roller coaster car immediately before it begins traveling down a long track
GrogVix [38]
At the top of the hill, the cars possess a large quantity of potential energy. Potential energy - the energy of vertical position - is dependent upon the mass of the object and the height of the object. The car's large quantity of potential energy is due to the fact that they are elevated to a large height above the ground. As the cars descend the first drop they lose much of this potential energy in accord with their loss of height. The cars subsequently gain kinetic energy. Kinetic energy - the energy of motion - is dependent upon the mass of the object and the speed of the object. The train of coaster cars speeds up as they lose height. Thus, their original potential energy (due to their large height) is transformed into kinetic energy (revealed by their high speeds). As the ride continues, the train of cars are continuously losing and gaining height. Each gain in height corresponds to the loss of speed as kinetic energy (due to speed) is transformed into potential energy (due to height). Each loss in height corresponds to a gain of speed as potential energy (due to height) is transformed into kinetic energy (due to speed). A roller coaster ride also illustrates the work and energy relationship. The work done by external forces is capable of changing the total amount of mechanical energy from an initial value to some final value. The amount of work done by the external forces upon the object is equal to the amount of change in the total mechanical energy of the object. The relationship is often stated in the form of the following mathematical equation.

KEinitial + PEinitial + Wexternal = KEfinal + PEfinal

The left side of the equation includes the total mechanical energy (KEinitial + PEinitial) for the initial state of the object plus the work done on the object by external forces (Wexternal) while the right side of the equation includes the total mechanical energy (KEfinal + PEfinal) for the final state of the object.

Once a roller coaster has reached its initial summit and begins its descent through loops, turns and smaller hills, the only forces acting upon the coaster cars are the force of gravity, the normal force and dissipative forces such as air resistance. The force of gravity is an internal force and thus any work done by it does not change the total mechanical energy of the train of cars. The normal force of the track pushing up on the cars is an external force. However, it is at all times directed perpendicular to the motion of the cars and thus is incapable of doing any work upon the train of cars. Finally, the air resistance force is capable of doing work upon the cars and thus draining a small amount of energy from the total mechanical energy which the cars possess. However, due to the complexity of this force and its small contribution to the large quantity of energy possessed by the cars, it is often neglected. By neglecting the influence of air resistance, it can be said that the total mechanical energy of the train of cars is conserved during the ride. That is to say, the total amount of mechanical energy (kinetic plus potential) possessed by the cars is the same throughout the ride. Energy is neither gained nor lost, only transformed from kinetic energy to potential energy and vice versa.

The conservation of mechanical energy by the coaster car in the above animation can be studied using a calculator. At each point in the ride, the kinetic and potential energies can be calculated using the following equations.

<span> KE = 0.5 * mass * (speed)^2 PE = mass * g * height</span>

If the acceleration of gravity value of 9.8 m/s/s is used along with an estimated mass of the coaster car (say 500 kg), the kinetic energy and potential energy and total mechanical energy can be determined

5 0
3 years ago
a 300-gram toy car and a 500-gram toy car are sitting on a shelf that is 2 meters highter than the floor. by what percent is the
andre [41]
We are given:

m1 = 300 grams
m2 = 500 grams
h1 = h2 = 2 meters

We have to determine the potential energy of both toy cars:

PE = mgh
PE1 = m1gh1
PE1 = 300 grams/1000g/kg * 9.81 m/s^2 * 2 meters
PE1 = 5.886 Joules

PE2 = m2gh2
PE2 = 500 grams/1000g/kg * 9.81 m/s^2 * 2 meters
PE2 = 9.81 Joules

PE2 - PE1 / PE2

9.81 - 5.886 Joules / 9.81 joules * 100 = 40.0%

Therefore, the 500-gram car is greater by 40% in terms of potential energy than the 300-gram car. This is because mass is directly proportional to the potential energy of an object. 

7 0
3 years ago
Read 2 more answers
A bicycle starts at 2.5m/s and accelerates along a straight path to a speed of 12.5m/s in a time of 4.5 seconds. What is the bic
zalisa [80]

Answer:

The bicyclist's acceleration is 2.2m/s^2

Explanation:

Given

u = 2.5m/s ---- Initial Velocity

v = 12.5m/s ---- Final Velocity

t = 4.5s ---- Time

Required

Determine the acceleration

This will be solved using the first equation of motion

v = u + at

Substitute values for v, u and a

12.5 = 2.5 + a * 4.5

12.5 = 2.5 + 4.5a

Collect Like Terms

4.5a = 12.5 - 2.5

4.5a = 10.0

Solve for a

a = 10.0/4.5

a = 2.2m/s^2 ---- (approximated)

<em>Hence, the bicyclist's acceleration is 2.2m/s^2</em>

3 0
3 years ago
A box having a mass of 0.2 kg is dragged across a horizontal floor by
Lyrx [107]

you can check attachment for answer.

kind regards

7 0
3 years ago
I'm walking 1.6m/s to 7-11 and it started to rain so I sped up to 2.7m/s in 1.2
olga nikolaevna [1]

Answer:

Explanation:

a = \frac{v_f-v_0}{t} which is the final velocity minus the initial velocity in the numerator, and the change in time in the denominator.  For us:

a=\frac{2.7-1.6}{1.2} so

a = .92 m/s/s (NOT negative because you're speeding up)

5 0
3 years ago
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