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Complete Question
A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.)
mm/s
Answer:
The drift velocity is 
Explanation:
From the question we are told that
The current on the copper is 
The cross-sectional area is 
The number of copper atom in the wire is mathematically evaluated
Where 
is the density of copper with a value 
is the Avogadro's number with a value 
Z is the molar mass of copper with a value 
So
Given the 1 atom is equivalent to 1 free electron then the number of free electron is
The current through the wire is mathematically represented as
substituting values
=>
Number a is a correct one
Answer:
a) The plasma membrane is called a selectively permeable membrane as it permits the movement of only certain molecules in and out of the cells. ... It allows hydrophobic molecules and small polar molecules diffuse through the lipid layer, but does not allow ions and large polar molecules cannot diffuse through the membrane
b) Plastids are present in the cells of plants. They are characterised by the presence of pigments. ... Chloroplasts contain chlorophyll and carotenoid pigments responsible for capturing the light energy that is necessary for photosynthesis. The chloroplasts are therefore known as the kitchen of the cell.
c) Lysosomes are known as the suicidal bag of the cell because it is capable of destroying its own cell in which it is present. It contains many hydrolytic enzymes which are responsible for the destruction process. This happens when either the cell is aged or gets infected by foreign agents like any bacteria or virus.
d) Mitochondria are often called the “powerhouses” or “energy factories” of a cell because they are responsible for making adenosine triphosphate (ATP), the cell's main energy-carrying molecule. ... In mitochondria, this process uses oxygen and produces carbon dioxide as a waste product.
e) In Hydra, the cells are arranged in two germinal layers—outer ectoderm and inner endoderm. Between these two layers is a layer of undifferentiated cells called mesoglea. Such kind of pattern of embryonic layers is seen in diploblastic animals. Hence, Hydra is a diploblastic animal.
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Answer:
-589.05 J
Explanation:
Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner
So, W = ΔK
W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)
So, substituting the values of the variables into the equation, we have
W = 1/2m(v₁² - v₀²)
W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)
W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)
W = 1/2 × 72.9 kg(-16.1604 m²/s²)
W = 1/2 × (-1178.09316 kgm²/s²)
W = -589.04658 kgm²/s²
W = -589.047 J
W ≅ -589.05 J
