The density of sample is 5 g/cm3
Given:
volume of sample = 20 cm3
mass of sample = 100 grams
To Find:
density of sample
Solution: Density is the measure of how much “stuff” is in a given amount of space. For example, a block of the heavier element lead (Pb) will be denser than the softer, lighter element gold (Au). A block of Styrofoam is less dense than a brick. It is defined as mass per unit volume
density = mass/volume
d = 100/20
d = 5 g/cm3
So, density of sample is 5 g/cm3
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Answer: reliable
Explanation:
Reliable (marketing research) information is collected from questions (measurements) that are free from systematic or statistical error. An absence of systematic error implies that the respondents (i.e., the sampled people) who answer questions actually understand what the questions were asking.
Answer:
(a) 3.24 w (b) 44.44 ohm
Explanation:
It is given that car draws 0.27 A current so current I = 0.27 A
The system has a voltage of 12 V
(a) Electrical power = voltage ×current 
(b) The resistance is defined as the ratio of voltage and current
So resistance 
Take a look at a simple reaction like the one below:
In this reaction some reactant A is turned into some product B. The rate of reaction can be represented by a decrease in concentration of A over time or as the increase of B over time. This is written:
The energy transfer in terms of work has the equation:
W = mΔ(PV)
To be consistent with units, let's convert them first as follows:
P₁ = 80 lbf/in² * (1 ft/12 in)² = 5/9 lbf/ft²
P₂ = 20 lbf/in² * (1 ft/12 in)² = 5/36 lbf/ft²
V₁ = 4 ft³/lbm
V₂ = 11 ft³/lbm
W = m(P₂V₂ - P₁V₁)
W = (14.5 lbm)[(5/36 lbf/ft²)(4 ft³/lbm) - (5/9 lbf/ft²)(11 lbm/ft³)]
W = -80.556 ft·lbf
In 1 Btu, there is 779 ft·lbf. Thus, work in Btu is:
W = -80.556 ft·lbf(1 Btu/779 ft·lbf)
<em>W = -0.1034 BTU</em>
