Answer:
The dynamic viscosity of the liquid is 0.727 kg/m*s
Explanation:
In the equation for that viscosimeter, ν = KR⁴t, <u>the terms K and R are not dependent on the liquid that is being tested</u>, unlike ν and t.
Using that equation and the data given in the problem, we can calculate the product of K and R⁴.
1.19*10⁻³m²/s = (KR⁴)* 1430 s
KR⁴=8,32*10⁻⁷m²/s²
We can now calculate the<em> </em><u><em>kinematic</em></u> viscosity of the unknown liquid.
ν=8,32*10⁻⁷m²/s²*900s
ν=7.49*10⁻⁴m²/s
The relationship between the <em>kinematic</em> viscosity and the <em>dynamic</em> viscosity is given by the equation μ=ν * ρ, where μ is the dynamic viscosity and ρ is the density. Thus:
μ=7.49*10⁻⁴m²/s * 970 kg/m³
μ=0.727 kg/m*s
It would be 50 because if its 10km/hour just multiply (10*5=50)
Hope it helps:)
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I believe the answer is the third option. Hope this helps! Please tell me if I am wrong or if there was an error in my answer... also sorry this answer is late.