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Characteristics of Solids
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Characteristics of solids (Grade 3 A and B)SCIENCE 3-Contents: Animals; Characteristics of Solids, Liquids, and Gases, and Sense Organs, Characteristics of Solid, Liquid and Gas QuizREVIEW GAME FOR CHARACTERISTICS OF SOLID AND LIQUIDIntroduction to Free Fall A free-falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. There are two important motion characteristics that are true of free-falling objects: • Free-falling objects do not encounter air resistance. • All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s for back-of-the-envelope calculations) Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a ticker tape trace or dot diagram of its motion would depict an acceleration. The dot diagram at the right depicts the acceleration of a free-falling object. The position of the object at regular time intervals - say, every 0.1 second - is shown. The fact that the distance that the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward. Recall from an earlier lesson, that if an object travels downward and speeds up, then its acceleration is downward. Free-fall acceleration is often witnessed in a physics classroom by means of an ever-popular strobe light demonstration. The room is darkened and a jug full of water is connected by a tube to a medicine dropper. The dropper drips water and the strobe illuminate the falling droplets at a regular rate - say once every 0.2 seconds. Instead of seeing a stream of water free-falling from the medicine dropper, several consecutive drops with increasing separation distance are seen. The pattern of drops resembles the dot diagram shown in the graphic at the right. The Acceleration of Gravity It was learned in the previous part of this lesson that a free-falling object is an object that is falling under the sole influence of gravity. A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it - the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s2. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. We will occasionally use the approximated value of 10 m/s2 in order to reduce the complexity of the many mathematical tasks that we will perform with this number. By so doing, we will be able to better focus on the conceptual nature of physics without too much of a sacrifice in numerical accuracy. g = 9.8 m/s2, downward Look It Up! Even on the surface of the Earth, there are local variations in the value of the acceleration of gravity (g). These variations are due to latitude, altitude and the local geological structure of the region. Recall from an earlier lesson that acceleration is the rate at which an object changes its velocity. It is the ratio of velocity change to time between any two points in an object's path. To accelerate at 9.8 m/s2 means to change the velocity by 9.8 m/s each second. If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern. Time (s) Velocity (m/s) 0 0 1 - 9.8 2 - 19.6 3 - 29.4 4 - 39.2 5 - 49.0 . Observe that the velocity-time data above reveal that the object's velocity is changing by 9.8 m/s each consecutive second. That is, the free-falling object has an acceleration of approximately 9.8 m/s2. Another way to represent this acceleration of 9.8 m/s2 is to add numbers to our dot diagram that we saw earlier in this lesson. The velocity of the ball is seen to increase as depicted in the diagram at the right. (NOTE: The diagram is not drawn to scale - in two seconds, the object would drop considerably further than the distance from shoulder to toes.) Representing Free Fall by Graphs • Early in Lesson 1 it was mentioned that there are a variety of means of describing the motion of objects. One such means of describing the motion of objects is through the use of graphs - position versus time and velocity vs. time graphs. In this part of Lesson 5, the motion of a free-falling motion will be represented using these two basic types of graphs. Representing Free Fall by Position-Time Graphs A position versus time graph for a free-falling object is shown below. Observe that the line on the graph curves. As learned earlier, a curved line on a position versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved. A further look at the position-time graph reveals that the object starts with a small velocity (slow) and finishes with a large velocity (fast). Since the slope of any position vs. time graph is the velocity of the object (as learned in Lesson 3), the small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Finally, the negative slope of the line indicates a negative (i.e., downward) velocity. Representing Free Fall by Velocity-Time Graphs A velocity versus time graph for a free-falling object is shown below. Observe that the line on the graph is a straight, diagonal line. As learned earlier, a diagonal line on a velocity versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward), it would be expected that its velocity-time graph would be diagonal. A further look at the velocity-time graph reveals that the object starts with a zero velocity (as read from the graph) and finishes with a large, negative velocity; that is, the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up is said to have a negative acceleration (if necessary, review the vector nature of acceleration). Since the slope of any velocity versus time graph is the acceleration of the object (as learned in Lesson 4), the constant, negative slope indicates a constant, negative acceleration. This analysis of the slope on the graph is consistent with the motion of a free-falling object - an object moving with a constant acceleration of 9.8 m/s/s in the downward direction. The Kinematic Equations The goal of this first unit has been to investigate the variety of means by which the motion of objects can be described. The variety of representations that we have investigated includes verbal representations, pictorial representations, numerical representations, and graphical representations (position-time graphs and velocity-time graphs). In Lesson 6, we will investigate the use of equations to describe and represent the motion of objects. These equations are known as kinematic equations. There are a variety of quantities associated with the motion of objects - displacement (and distance), velocity (and speed), acceleration, and time. Knowledge of each of these quantities provides descriptive information about an object's motion. For example, if a car is known to move with a constant velocity of 22.0 m/s, North for 12.0 seconds for a northward displacement of 264 meters, then the motion of the car is fully described. And if a second car is known to accelerate from a rest position with an eastward acceleration of 3.0 m/s2 for a time of 8.0 seconds, providing a final velocity of 24 m/s, East and an eastward displacement of 96 meters, then the motion of this car is fully described. These two statements provide a complete description of the motion of an object. However, such completeness is not always known. It is often the case that only a few parameters of an object's motion are known, while the rest are unknown. For example as you approach the stoplight, you might know that your car has a velocity of 22 m/s, East and is capable of a skidding acceleration of 8.0 m/s2, West. However you do not know the displacement that your car would experience if you were to slam on your brakes and skid to a stop; and you do not know the time required to skid to a stop. In such an instance as this, the unknown parameters can be determined using physics principles and mathematical equations (the kinematic equations). The BIG 4 The kinematic equations are a set of four equations that can be utilized to predict unknown information about an object's motion if other information is known. The equations can be utilized for any motion that can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant acceleration motion. They can never be used over any time period during which the acceleration is changing. Each of the kinematic equations include four variables. If the values of three of the four variables are known, then the value of the fourth variable can be calculated. In this manner, the kinematic equations provide a useful means of predicting information about an object's motion if other information is known. For example, if the acceleration value and the initial and final velocity values of a skidding car is known, then the displacement of the car and the time can be predicted using the kinematic equations. Lesson 6 of this unit will focus upon the use of the kinematic equations to predict the numerical values of unknown quantities for an object's motion. The four kinematic equations that describe an object's motion are: There are a variety of symbols used in the above equations. Each symbol has its own specific meaning. The symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Each of these four equations appropriately describes the mathematical relationship between the parameters of an object's motion. As such, they can be used to predict unknown information about an object's motion if other information is known. In the next part of Lesson 6 we will investigate the process of doing this. Kinematic Equations and Problem-Solving The four kinematic equations that describe the mathematical relationship between the parameters that describe an object's motion were introduced in the previous part of Lesson 6. The four kinematic equations are: In the above equations, the symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stand for the acceleration of the object. And the symbol v stands for the instantaneous velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Problem-Solving Strategy In this part of Lesson 6 we will investigate the process of using the equations to determine unknown information about an object's motion. The process involves the use of a problem-solving strategy that will be used throughout the course. The strategy involves the following steps: 1. Construct an informative diagram of the physical situation. 2. Identify and list the given information in variable form. 3. Identify and list the unknown information in variable form. 4. Identify and list the equation that will be used to determine unknown information from known information. 5. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information. 6. Check your answer to ensure that it is reasonable and mathematically correct. The use of this problem-solving strategy in the solution of the following problem is modeled in Examples A and B below. Example Problem A . Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.) The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. Note that the vf value can be inferred to be 0 m/s since Ima's car comes to a stop. The initial velocity (vi) of the car is +30.0 m/s since this is the velocity at the beginning of the motion (the skidding motion). And the acceleration (a) of the car is given as - 8.00 m/s2. (Always pay careful attention to the + and - signs for the given quantities.) The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = +30.0 m/s vf = 0 m/s a = - 8.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vf, vi, a, and d. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (30.0 m/s)2 + 2 • (-8.00 m/s2) • d 0 m2/s2 = 900 m2/s2 + (-16.0 m/s2) • d (16.0 m/s2) • d = 900 m2/s2 - 0 m2/s2 (16.0 m/s2)*d = 900 m2/s2 d = (900 m2/s2)/ (16.0 m/s2) d = (900 m2/s2)/ (16.0 m/s2) d = 56.3 m The solution above reveals that the car will skid a distance of 56.3 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. It takes a car a considerable distance to skid from 30.0 m/s (approximately 65 mi/hr) to a stop. The calculated distance is approximately one-half a football field, making this a very reasonable skidding distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step of the strategy involves the identification and listing of known information in variable form. Note that the vi value can be inferred to be 0 m/s since Ben's car is initially at rest. The acceleration (a) of the car is 6.00 m/s2. And the time (t) is given as 4.10 s. The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0 m/s t = 4.10 s a = 6.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are t, vi, a, and d. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. d = (0 m/s) • (4.1 s) + ½ • (6.00 m/s2) • (4.10 s)2 d = (0 m) + ½ • (6.00 m/s2) • (16.81 s2) d = 0 m + 50.43 m d = 50.4 m The solution above reveals that the car will travel a distance of 50.4 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. A car with an acceleration of 6.00 m/s/s will reach a speed of approximately 24 m/s (approximately 50 mi/hr) in 4.10 s. The distance over which such a car would be displaced during this time period would be approximately one-half a football field, making this a very reasonable distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! The two example problems above illustrate how the kinematic equations can be combined with a simple problem-solving strategy to predict unknown motion parameters for a moving object. Provided that three motion parameters are known, any of the remaining values can be determined. In the next part of Lesson 6, we will see how this strategy can be applied to free fall situations. Or if interested, you can try some practice problems and check your answer against the given solutions. Kinematic Equations and Free Fall As mentioned in Lesson 5, a free-falling object is an object that is falling under the sole influence of gravity. That is to say that any object that is moving and being acted upon only be the force of gravity is said to be "in a state of free fall." Such an object will experience a downward acceleration of 9.8 m/s/s. Whether the object is falling downward or rising upward towards its peak, if it is under the sole influence of gravity, then its acceleration value is 9.8 m/s/s. Like any moving object, the motion of an object in free fall can be described by four kinematic equations. The kinematic equations that describe any object's motion are: The symbols in the above equation have a specific meaning: the symbol d stands for the displacement; the symbol t stands for the time; the symbol a stands for the acceleration of the object; the symbol vi stands for the initial velocity value; and the symbol vf stands for the final velocity. Applying Free Fall Concepts to Problem-Solving There are a few conceptual characteristics of free fall motion that will be of value when using the equations to analyze free fall motion. These concepts are described as follows: • An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object. • If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity (vf) after traveling to the peak would be assigned a value of 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height. These four principles and the four kinematic equations can be combined to solve problems involving the motion of free-falling objects. The two examples below illustrate application of free fall principles to kinematic problem-solving. In each example, the problem solving strategy that was introduced earlier in this lesson will be utilized. Example Problem A Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground. The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 8.52 meters. The displacement (d) of the shingles is -8.52 m. (The - sign indicates that the displacement is downward). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. For example, the vi value can be inferred to be 0 m/s since the shingles are dropped (released from rest; see note above). And the acceleration (a) of the shingles can be inferred to be -9.8 m/s2 since the shingles are free-falling (see note above). (Always pay careful attention to the + and - signs for the given quantities.) The next step of the solution involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the time of fall. So t is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0.0 m/s d = -8.52 m a = - 9.8 m/s2 t = ?? The next step involves identifying a kinematic equation that allows you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are d, vi, a, and t. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. -8.52 m = (0 m/s) • (t) + ½ • (-9.8 m/s2) • (t)2 -8.52 m = (0 m) *(t) + (-4.9 m/s2) • (t)2 -8.52 m = (-4.9 m/s2) • (t)2 (-8.52 m)/(-4.9 m/s2) = t2 1.739 s2 = t2 t = 1.32 s The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable. The calculated time easily falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 26.2 m/s. The initial velocity (vi) of the vase is +26.2 m/s. (The + sign indicates that the initial velocity is an upwards velocity). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. Note that the vf value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory (see note above). The acceleration (a) of the vase is -9.8 m/s2 (see note above). The next step involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the vase (the height to which it rises above its starting height). So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 26.2 m/s vf = 0 m/s a = -9.8 m/s2 d = ?? The next step involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vi, vf, a, and d. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (26.2 m/s)2 + 2 •(-9.8m/s2) •d 0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2) •d (-19.6 m/s2) • d = 0 m2/s2 -686.44 m2/s2 (-19.6 m/s2) • d = -686.44 m2/s2 d = (-686.44 m2/s2)/ (-19.6 m/s2) d = 35.0 m The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never make it further than one football field in height (approximately 100 m), yet will surely make it past the 10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! Kinematic equations provide a useful means of determining the value of an unknown motion parameter if three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And in many cases, another motion parameter can be inferred through a solid knowledge of some basic kinematic principles.Community Engagement, Solidarity and Citizenship Lesson 1 Definition and Characteristics of CommunityMake mcq quiz with 4 option in which one is correct -'10 Basis of Material Science • .....;;;";;;"~~;;,,;;,,,,;.;.,,;;,,,;,,;.;,.,------------ 6. Temporary materials: Some materials are meant to be placed in the oral cavity for a short period of time for different reasons. • Temporary crowns: While a permanent crown is prepared in the dental laboratory, the patient must wait for few days before it can be fabricated and cemented into place. Does patient experience any problems during this time period? If the tooth is vital (the pulp is alive), the patient is likely to experience pain and sensitivity while eating and drinking, also it looks unesthetic. What can be done to solve this problem? A temporary crown is placed before the patient leaves the clinic. It is constructed and luted in the same appointment in which the crown preparation is done. Temporary crowns are not very strong or esthetic but they serve adequately till the permanent crown is ready to be cemented. • Temporary restorations: Sometimes it is difficult to decide immediately the best line of treatment for a particular tooth. The exact condition of the pulp may not be obvious to the dentist from the patient's symptoms. A dentist removes all or part of the decay and then places a temporary restoration to have time to observe the behaviour of the pulp or to give the pilip time to heal before deciding the further treatment required. Classification based on Location of Fabrication 4,9 Materials can be classified based on the location of fabrication into: • Direct restorative materials. • Indirect restorative materials Direct restorative materials: They include those materials which are used to restore cavity preparations directly in the oral cavity (Box 1.5). Box 1.5: Examples of direct restorative materials Amalgam, composites, glass ionomer and other materials, which set by chemical reactions in the mouth. Indirect restorative materials: It includes those restorations which must be fabricated outside the mouth, indirectly on a cast/ model/ die, because their processing condition would harm oral tissues. Materials used in the construction of such prosthesis are called indirect restorative materials (Box 1.6). Box 1.6: Examples of indirect restorative materials Gold inlays, crowns of metal, ceramic and polymers, which are processed at elevated temperatures. Some indirect composite restorations can be processed under specific wavelength of light, e.g. Ceramage. Classification based on Longevity of Use 1. Permanent restorations: These restorations are not planned to be replaced for a particular time period. Though they are referred to as permanent, actually they are not, e.g. fillings, crowns, bridges and dentures do not last forever (Fig. 1.5). 2. Temporary restorations: These restorations are planned to be replaced in a short period of time, such as few days to weeks. For ~ Permanent C/) c c -.2 0 c- :;::; Cll co Interim ~ Q; 0 .8ll::1iJ C/) o~ Cll a:: c:=:J Temporary Time period Fig. 1.5: Diagram depicting the time period of use of a restoration. (Arrow in permanent restoration depicts that such restorations are not planned to be replaced for a long period of time.) Introducton to Dental Materials Dental materials Box 1.7: Characteristics of metals 1. High thermal and electrical conductivity 2. Ductility (pure metals are very soft and they can be bent without breaking) 3. Opacity (they do not transmit light) 4. Luster (they have a surface that strongly reflects light and appears bright and shiny) 5. They tend to dissolve to some extent in water or other aqueous solutions, producing cations. 6. All metals are white (actually gray) except for gold, which is yellow, and copper, which is reddish. 7. All metals are solid at room temperature except mercury, which is liquid at room temperature and is used with silver alloys as amalgam. 8. All metals have high melting temperatures because of high strength of the metallic bond that holds the atoms together. 3. Polymers 4. Composites Composites are mixtures of two or more of the first three classes in which the different components remain distinct from one another in the final structure. A common example is composite resin. Fig. 1.7a: Three-dimensional structure of iron (metal) Metals Metals are the oldest of the three classes of materials that have been used as dental materials. Metals are characterized by metallic bonds (Box 1.7) which will be discussed in the next chapter. Metals solidify with their atoms in a regular or crystalline arrangement (see Chapter 2), often in the form of a cube (Fig. 1.7a). example, temporary fillings done in a tooth during root canal treatment, which have to be replaced within 2-4 days during subsequent visits. They are used to protect the tooth and provide function till the final restoration is done. 3. Interim restoration: At times, dental treatment requires "long-term" definite temporary restorations or "interim" restorations. For examle, a 7-year-old child, met with trauma and fractured one of his central incisors. A large composite build- up may serve his immediate requirement until the root formation is completed and a permanent crown is placed. 5 Classification based on the Chemical Nature of the Material These are the atoms that make up a material and the way they are bonded together determine the properties of that materiaLS Weak bonds make for weak materials and vice versa (Table 1.4). Materials can be classified into different categories based on their primary atomic bonds (Fig. 1.6): 1. Metals 2. Ceramics Fig. 1.6: Classification of dental materials based on chemical nature 12 Basis of Material Science Box 1.9: Benefits of ceramics in dentistry 1. Many ceramic oxides are used as pigmenting agents. These oxides produce good range of colors. Due to this characteristic, we are able to match almost any tooth color with good esthetic results. 2. They are inert, i.e. not chemically reactive. This quality provides ceramics with good bio- compatibility. 3. Ceramic materials are translucent, like natural teeth. This translucency gives the ceramic crown a more natural appearance than any other dental material. Fig. 1.7b: Internal arrangement of tetrahedral structure of ceramic (silica) four large oxygen atoms surround smaller silicon atom Ceramics A ceramic is a compound formed by the union of a metallic and a non-metallic element (Box 1.8). Most of these materials are oxides, formed by the union of oxygen with metals such as silicon, aluminum, calcium and magnesium (Fig.1.7b). Ceramics may be simple or complex. Examples of simple ceramics are alumina and silica. Examples of complex ceramics are feldspar (potassium aluminum silicate) and kaolin (hydrated aluminum silicate). Ceramics may be crystalline or non- crystalline (i.e. amorphous). Porcelain is a specific type of ceramic used extensively in dentistry (Box 1.9). Box 1.8: Characteristics of ceramics 1. High melting points. 2. Brittleness, which means they cannot be bent or deformed (no sliding) to any extent without actually cracking and breaking. 3. They are poor conductor of heat and electricity. 4. They are chemically inert. 5. They have excellent esthetic result in terms of matching natural teeth. Fig. 1.8: Stucture of synthetic polymer Polymers They are the latest addition (early to mid- 1900s) to dental materials. Most of the polymers are nowadays synthesized by humans. Polymers are giant, long-chain organic molecules (Fig. 1.8). Polymers are characterized by covalent bonds within each molecule, giving them tremendous strength in a single direction. Try to break a nylon rope by pulling it! They are poor conductors of heat and electri- city. Most polymers have a structure containing thousands of carbon atoms linked together like beads on a string. Others, such as silicone polymers are formed with silicon-oxygen bonds. Introducton to Dental Materials Table 1.4: Characteristics of different materials 13 Characteristics Bond Properties Crystal structure Metals Metallic bonding High strength and hardness, high electrical and thermal conductivity BCC, FCC, or HCP unit cells Ceramics Ionic or covalent bonding, or both High hardness and stiffness, electrically insulating, refractory, and chemically inert Crystalline or amorphous Polymers Covalent bonding Low sensitivity, high electrical resistivity, and low thermal conductivity, strength and stiffness vary widely Amorphous and crystalline Composites Composites are combinations of any of the basic ceramic, metallic and polymeric materials (Box 1.10). Each material that makes up composites is called a phase. Their properties tend to be somewhere between those of their basic constituents and are used to enhance their performance, longevity and handling chracterstics. Box 1.10: Types of composites in dentistry 1. Ceramic - metallic composite: Tungsten carbide bur. 2. Metal - polymer composite: Die materials in dental laboratory. 3. Ceramic - polymer composite: Enamel, dentin, bone and restorative composites. A composite is a kind of "combination" of materials, which compliment each other. The properties lacking in one material are compensated by those of the other material. For example, restorative composite has two phases, namely resin and fillers. Teeth and bones are examples of natural composites. Enamel is a composite of hydroxyapatite (which is a ceramic material) and protein (which is a polymer). EVALUATION OF DENTAL MATERIALS Most manufacturers of dental materials maintain a quality assurance programme (As per international standard like ADA specifications) and materials are thoroughly tested before being released into the market for dental practitioner (Fig. 1.9). Laboratory Evaluations Most ADA/ ANSI specifications involve laboratory tests. The tests performed as per these specifications are useful but they all are performed in vitro, (carried out in the laboratory away from the clinical conditions) which have a lot of limitations in clinical practice.lO Clinical Notes 1. For example, most of the direct restorative materials are tested for their compressive strength but ultimately the material is subjected to a combination of compressive, tensile and shear stresses, which may decide the final success or failure of the material under masticatory load. 2. Similarly upper dentures mostly fracture along the midline because of bending. Hence a bending or transverse strength ~B-a-s-is-o-f-M-a-t-e-ria-I-S~c-ie-n-c-e-------------- ---------. test is far more meaningful for denture base materials than a compression test. Clinical Trials The majority of new materials are subjected to extensive clinical trials normally in co-operation with a dental college or hospital departments prior to their release. CONCLUSION As the number of available materials is going up, it is important that the dentist remains more aware about new products so that their judgement about the selection of material remains successful. Materials which have not been thoroughly evaluated should be avoided, specially with clinical dentistry falling under Consumer Protection Act (CPA). I Research and development I iI Manufacturer/analysis Ideal requirements for clinical use: Thermal, optical, mechanical, chemical, biological Available materials and their properties are evaluated Launch of new I product Choice and selection of material by the dentist Critical assessment based on clinical performance I I H feedback to IThe plasma membrane (also called the cell membrane) has several functions. For example, it allows only certain molecules to enter or leave the cell. It separates internal metabolic reactions from the external environment. In addition, the plasma membrane allows the cell to excrete wastes and to interact with its environment. Membrane Lipids The plasma membrane, as well as the membranes of cell organelles, is made primarily of phospholipids. Phospholipids have a polar, hydrophilic (“water-loving”) phosphate head and two nonpolar, hydrophobic (“water-fearing”) fatty acid tails. Water molecules sur- round the plasma membrane. The phospholipids line up so that their heads point outward toward the water and their tails point inward, away from water. The result is a double layer called a phospholipid bilayer, as shown in Figure 4-10. The cell membranes of eukaryotes also contain lipids, called sterols, between the tails of the phospho- lipids. The major membrane sterol in animal cells is cholesterol. Sterols in the plasma membrane make the membrane more firm and prevent the membrane from freezing at low temperatures. SECTION 3 OBJECTIVES ● Describe the structure and function of a cell’s plasma membrane. ● Summarize the role of the nucleus. ● List the major organelles found in the cytosol, and describe their roles. ● Identify the characteristics of mitochondria. ● Describe the structure and function of the cytoskeleton. VOCABULARY phospholipid bilayer chromosome nuclear envelope nucleolus ribosome mitochondrion endoplasmic reticulum Golgi apparatus lysosome cytoskeleton microtubule microfilament cilium flagellum centriole Cell membranes are made of a phospholipid bilayer. Each phospholipid molecule has a polar “head” and a two-part nonpolar “tail.” FIGURE 4-10 Copyright © by Holt, Rinehart and Winston. All rights reserved. 78 CHAPTER 4 OUTSIDE OF CELL INSIDE OF CELL 1. Cell-surface marker: Glycoprotein that identifies cell type 3. Enzyme: Assists chemical reactions inside the cell 2. Receptor protein: Recognizes and binds to substances outside the cell 4. Transport protein: Helps substances move across cell membrane Carbohydrate portion Protein portion Phospholipid heads Phospholipid tails Phospholipid Cholesterol bilayer Membrane Proteins Plasma membranes often contain specific proteins embedded within the lipid bilayer. These proteins are called integral proteins. Figure 4-11 shows that some integral proteins, such as cell surface markers, emerge from only one side of the membrane. Others, such as receptor proteins and transport proteins, extend across the plasma membrane and are exposed to both the cell’s interior and exterior environments. Proteins that extend across the plasma membrane are able to detect environmental signals and transmit them to the inside of the cell. Peripheral proteins, such as the enzyme shown in Figure 4-11, lie on only one side of the membrane and are not embedded in it. As Figure 4-11 shows, integral proteins exposed to the cell’s external environment often have carbohydrates attached. These carbohydrates can act as labels on cell surfaces. Some labels help cells recognize each other and stick together. Viruses can use these labels as docks for entering and infecting cells. Integral proteins play important roles in actively transporting molecules into the cell. Some act as channels or pores that allow certain substances to pass. Other integral proteins bind to a mol- ecule on the outside of the cell and then transport it through the membrane. Still others act as sites where chemical messengers such as hormones can attach. Fluid Mosaic Model A cell’s plasma membrane is surprisingly dynamic. Scientists describe the cell membrane as a fluid mosaic. The fluid mosaic model states that the phospholipid bilayer behaves like a fluid more than it behaves like a solid. The membrane’s lipids and pro- teins can move laterally within the bilayer, like a boat on the ocean. As a result of such lateral movement, the pattern, or “mosaic,” of lipids and proteins in the cell membrane constantly changes.