Answer:
here are four basic cost behavior patterns: fixed, variable, mixed (semivariable), and step which graphically would appear as below. The relevant range is the range of production or sales volume over which the assumptions about cost behavior are valid. Often, we describe them as time-related costs.
Explanation:
The general types of cost behavior fall into three categories. First is variable costs, which vary directly with changes in business activity. For example, there is a specific direct materials cost associated with each product sold. Second is fixed costs, which do not change in response to business activity levels.
200 calories each day-decreasing your caloric intake would remove 2100 calories a week, with 1400 unaccounted for. Therefore, the remaining calories could be split up equally between the seven days, with 200 burned each day
Single-celled organisms exchange gases directly across their cell membrane. However, the slow diffusion rate of oxygen relative to carbon dioxide limits the size of single-celled organisms. Simple animals that lack specialized exchange surfaces have flattened, tubular, or thin shaped body plans, which are the most efficient for gas exchange. However, these simple animals are rather small in size.
Frozen water is less denser than liquid water so that's why ice floats on the water in cold areas.
<h3>Is frozen water more or less dense than liquid water?</h3>
Ice is less dense than liquid water because when the water becomes solid due to low temperature, the density of ice become decreases. Due to less density, the ice floats on the water surface.
Water is attracted due to its polar nature so adhesive forces pull the water toward other molecules. Water is transported in plants through both cohesive and adhesive forces. These forces pull water and the dissolved minerals from the roots to the upper parts of the plant.
So we can conclude that frozen water is less denser than liquid water so that's why ice floats on the water in cold areas.
Learn more about density here: brainly.com/question/1354972
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Answer:
Q = 5227.5 J
Explanation:
Given data:
Mass of water = 50.0 g
Temperature increase by = ΔT = 25°C
Specific heat capacity of water = 4182 J/Kg.°C
Heat required = ?
Solution:
(50.0 g×1 Kg/1000 g=0.05 Kg)
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Q = 0.05 Kg ×4182 J/Kg.°C × 25°C
Q = 5227.5 J
