Answer:
Explanation:
1.3 PHYSICAL AND CHEMICAL PROPERTIES
Learning Objectives
By the end of this section, you will be able to:
Identify properties of and changes in matter as physical or chemical
Identify properties of matter as extensive or intensive
The characteristics that enable us to distinguish one substance from another are called properties. A physical property is a characteristic of matter that is not associated with a change in its chemical composition. Familiar examples of physical properties include density, color, hardness, melting and boiling points, and electrical conductivity. We can observe some physical properties, such as density and color, without changing the physical state of the matter observed. Other physical properties, such as the melting temperature of iron or the freezing temperature of water, can only be observed as matter undergoes a physical change. A physical change is a change in the state or properties of matter without any accompanying change in its chemical composition (the identities of the substances contained in the matter). We observe a physical change when wax melts, when sugar dissolves in coffee, and when steam condenses into liquid water (Figure 1). Other examples of physical changes include magnetizing and demagnetizing metals (as is done with common antitheft security tags) and grinding solids into powders (which can sometimes yield noticeable changes in color). In each of these examples, there is a change in the physical state, form, or properties of the substance, but no change in its chemical composition.
Answer:
critical stress required for the propagation is 27.396615 × N/m²
Explanation:
given data
specific surface energy = 0.90 J/m²
modulus of elasticity E = 393 GPa = 393 × N/m²
internal crack length = 0.6 mm
to find out
critical stress required for the propagation
solution
we will apply here critical stress formula for propagation of internal crack
( σc ) = .....................1
here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm = 0.3 × m
so now put value in equation 1 we get
( σc ) =
( σc ) =
( σc ) = 27.396615 × N/m²
so critical stress required for the propagation is 27.396615 × N/m²
Answer: False
Explanation: Sandwich materials are usually in composite material form which has a fabrication of two thin layers which are stiff in nature and have light weighing and thick core .The construction is based on the ratio that is of stiffness to the weight .Therefore, the density of the material in the core is not high and are only connected with the skin layer through adhesive .So the given statement is false that sandwich materials typically use a high density core with non- structural cover plates.
Question:
The following tabulated data were gathered from a series of Charpy impact tests on a commercial low-carbon steel alloy.
Temperature(∘C)50403020100-10-20-30-40Impact energy (J)76767158382314951.5
(a) Plot the data as impact energy versus temperature.
(b) Determine a ductile-to-brittle transition temperature as the temperature corresponding to the average of the maximum and minimum impact energies.
(c) Determine a ductile-to-brittle transition temperature as the temperature at which the impact energy is 20 J.
Answer:
a) see the attached graph
b) max E = 76 and Min E =1.53 average = 77.5/2 = 38.75 J
this corresponds to about 10° C
c ) at E = 20 J temperature is about -2°C
Answer:
filter drier
Explanation:
Any refrigeration or air conditioning system needs filter driers as a crucial part. In a refrigeration or air conditioning system, a filter drier serves two crucial purposes: first, it adsorbs system impurities like water, which can produce acids, and second, it provides physical filtering. To guarantee optimal and cost-effective dryer design, each component must be evaluated. The most crucial role of a dryer is to be able to drain water from a refrigeration system. Water can originate from a variety of places, including motor windings, equipment leaks, retained air from poor drainage, and others.