Greetings!
The correct answer choice is Choice 4.
<em>Why?</em>
In a scientific experiment the only thing being changed is the independent variable. Everything else should stay the same.
In this experiment, the independent variable is the amount of sunlight each plant should receive. <em>Here's a tip</em>- when looking for and independent variable look for whats being changed on purpose.
Hope this helps!
~Fluerie
Answer:
a) The shear stress is 0.012
b) The shear stress is 0.0082
c) The total friction drag is 0.329 lbf
Explanation:
Given by the problem:
Length y plate = 2 ft
Width y plate = 10 ft
p = density = 1.938 slug/ft³
v = kinematic viscosity = 1.217x10⁻⁵ft²/s
Absolute viscosity = 2.359x10⁻⁵lbfs/ft²
a) The Reynold number is equal to:
The boundary layer thickness is equal to:
ft
The shear stress is equal to:
b) If the railing edge is 2 ft, the Reynold number is:
The boundary layer is equal to:
The sear stress is equal to:
c) The drag coefficient is equal to:
The friction drag is equal to:
No. the answer to the question if can an argon atom undergo vibrational motion is no. it can not even spin either. the argon atom, or the argon is a chemical element that is the third most abundant gas in the earth's atmosphere. it is ore than twice as abundance as water vapor. Thank you for this question.
This question is personal preference... just answer with whatever YOU think, there probably isn’t s wrong answer if I’d guess.
Answer:
a. 0.342 kg-m² b. 2.0728 kg-m²
Explanation:
a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m
I = 1/2MR²
= 1/2 × 56.5 kg × (0.11 m)²
= 0.342 kgm²
So the moment of inertia of the skater is
b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)
I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m
I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²
= 0.1802 kg-m² + 0.6852 kg-m²
= 0.8654 kg-m²
The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².
So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²
