Answer:
1.2727 stokes
Explanation:
specific gravity of fluid A = 1.65
Dynamic viscosity = 210 centipoise
<u>Calculate the kinematic viscosity of Fluid A </u>
First step : determine the density of fluid A
Pa = Pw * Specific gravity = 1000 * 1.65 = 1650 kg/m^3
next : convert dynamic viscosity to kg/m-s
210 centipoise = 0.21 kg/m-s
Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A
= 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec
Convert to stokes = 1.2727 stokes
Answer:
Among the different types of excavation protection system, as a way of preventing accidents against cave-ins, the sloping involves cutting back the trench wall at an angle inclined away from the excavation. Shoring requires installing aluminum hydraulic or other types of supports to prevent soil movement and cave-ins. Shielding protects workers by using trench boxes or other types of supports to prevent soil cave-ins (OSHA). In addition, the regulations do not allow employees to work on excavations where there is an accumulation of water. If this occurs, water on the site must be constantly removed by suitable equipment preventing water from accumulating. The entry of surface water into the excavations must also be prevented by means of diversion ditches, dam, or other suitable means.
Explanation:
Answer:
Headlights are required to be used 1/2 hour after sunset to 1/2 hour before sunrise, when windshield wipers are being used, when visibility is less than 1000 feet, or when there is insufficient light or adverse weather.
Explanation:
hope this helps
Answer:
t = 25.10 sec
Explanation:
we know that Avrami equation

here Y is percentage of completion of reaction = 50%
t is duration of reaction = 146 sec
so,


taking natural log on both side
ln(0.5) = -k(306.6)

for 86 % completion




t = 25.10 sec
Answer:
a) at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
b) daylight (d) = 0.50 μm
Incandescent ( i ) = 1 μm
Explanation:
To Calculate the band emission fractions we will apply the Wien's displacement Law
The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as
F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )
<em>Values are gotten from the table named: blackbody radiati</em>on functions
<u>a) Calculate the band emission fractions for the visible region</u>
at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
attached below is a detailed solution to the problem
<u>b)calculate wavelength corresponding to the maximum spectral intensity</u>
For daylight ( d ) = 2898 μm *k / 5800 k = 0.50 μm
For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm