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The terminal velocity of a sphere of radius 25 mm and specific weight 27 kN/m3 traveling downwards in a liquid of viscosity 0.00089 Pas and specific weight 10 kN/m3 is __________ m/s. (Answer up to two decimal places)
Correct answer is '2652.93'. Can you explain this answer?
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The terminal velocity of a sphere of radius 25 mm and specific weight...
Terminal velocity
Terminal velocity is the maximum velocity that an object can attain when falling through a fluid. It occurs when the drag force acting on the object is equal to the gravitational force pulling it downwards.
Formula for terminal velocity
The formula for terminal velocity is given by:
V_t = (2 * (ρ_s - ρ_f) * g * r^2) / (9 * η)
Where:
V_t = Terminal velocity
ρ_s = Specific weight of the sphere
ρ_f = Specific weight of the fluid
g = Acceleration due to gravity
r = Radius of the sphere
η = Viscosity of the fluid
Given values
ρ_s = 27 kN/m^3 = 27,000 N/m^3
ρ_f = 10 kN/m^3 = 10,000 N/m^3
r = 25 mm = 0.025 m
η = 0.00089 Pas
Calculating terminal velocity
Substituting the given values into the formula, we have:
V_t = (2 * (27,000 - 10,000) * 9.8 * (0.025)^2) / (9 * 0.00089)
Simplifying the equation, we get:
V_t = (2 * 17,000 * 9.8 * 0.000625) / 0.00801
V_t = 212,150 / 0.00801
V_t ≈ 26,506.87 m/s
Rounding to two decimal places, the terminal velocity is approximately 26,506.87 m/s.
Terminal velocity is the maximum velocity that an object can attain when falling through a fluid. It occurs when the drag force acting on the object is equal to the gravitational force pulling it downwards.
Formula for terminal velocity
The formula for terminal velocity is given by:
V_t = (2 * (ρ_s - ρ_f) * g * r^2) / (9 * η)
Where:
V_t = Terminal velocity
ρ_s = Specific weight of the sphere
ρ_f = Specific weight of the fluid
g = Acceleration due to gravity
r = Radius of the sphere
η = Viscosity of the fluid
Given values
ρ_s = 27 kN/m^3 = 27,000 N/m^3
ρ_f = 10 kN/m^3 = 10,000 N/m^3
r = 25 mm = 0.025 m
η = 0.00089 Pas
Calculating terminal velocity
Substituting the given values into the formula, we have:
V_t = (2 * (27,000 - 10,000) * 9.8 * (0.025)^2) / (9 * 0.00089)
Simplifying the equation, we get:
V_t = (2 * 17,000 * 9.8 * 0.000625) / 0.00801
V_t = 212,150 / 0.00801
V_t ≈ 26,506.87 m/s
Rounding to two decimal places, the terminal velocity is approximately 26,506.87 m/s.
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Community Answer
The terminal velocity of a sphere of radius 25 mm and specific weight...
Given: r = 25 mm = 0.025 m, Ws = 27 kN/m3 = 27,000 N/m3,
μ = 0.00089 Pas and W = 10 kN/m3 = 10000 N/m3
Terminal velocity of sphere is
= 2652.93 m/s
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The terminal velocity of a sphere of radius 25 mm and specific weight 27 kN/m3 traveling downwards in a liquid of viscosity 0.00089 Pas and specific weight 10 kN/m3 is __________ m/s. (Answer up to two decimal places)Correct answer is '2652.93'. Can you explain this answer?
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The terminal velocity of a sphere of radius 25 mm and specific weight 27 kN/m3 traveling downwards in a liquid of viscosity 0.00089 Pas and specific weight 10 kN/m3 is __________ m/s. (Answer up to two decimal places)Correct answer is '2652.93'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The terminal velocity of a sphere of radius 25 mm and specific weight 27 kN/m3 traveling downwards in a liquid of viscosity 0.00089 Pas and specific weight 10 kN/m3 is __________ m/s. (Answer up to two decimal places)Correct answer is '2652.93'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The terminal velocity of a sphere of radius 25 mm and specific weight 27 kN/m3 traveling downwards in a liquid of viscosity 0.00089 Pas and specific weight 10 kN/m3 is __________ m/s. (Answer up to two decimal places)Correct answer is '2652.93'. Can you explain this answer?.
The terminal velocity of a sphere of radius 25 mm and specific weight 27 kN/m3 traveling downwards in a liquid of viscosity 0.00089 Pas and specific weight 10 kN/m3 is __________ m/s. (Answer up to two decimal places)Correct answer is '2652.93'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The terminal velocity of a sphere of radius 25 mm and specific weight 27 kN/m3 traveling downwards in a liquid of viscosity 0.00089 Pas and specific weight 10 kN/m3 is __________ m/s. (Answer up to two decimal places)Correct answer is '2652.93'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The terminal velocity of a sphere of radius 25 mm and specific weight 27 kN/m3 traveling downwards in a liquid of viscosity 0.00089 Pas and specific weight 10 kN/m3 is __________ m/s. (Answer up to two decimal places)Correct answer is '2652.93'. Can you explain this answer?.
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