Answer:
The pressure drop is 269.7N/m^2
Explanation:
∆P = ∆h × rho × g
∆h = 3.2cm = 3.2/100 = 0.032m, rho = 860kg/m^3, g = 9.8m/s^2
∆P = 0.032×860×9.8 = 269.7N/m^2
Answer: the absolute static pressure in the gas cylinder is 82.23596 kPa
Explanation:
Given that;
patm = 79 kPa, h = 13 in of H₂O,
A sketch of the problem is uploaded along this answer.
Now
pA = patm + 13 in of H₂O ( h × density × g )
pA= 79 + (13 × 0.0254 × 9.8 × 1000/1000)
pA = 82.23596 kPa
the absolute static pressure in the gas cylinder is 82.23596 kPa
Answer:
Effective reach and Frequency
Explanation:
Effective Reach is percentage of target audience that is exposed to a particular ad and receives given message to affect sales and purchase who are reached at or above effective frequency level. Here effective frequency level is the number of exposures necessary to make an impact and attain communication goal.
Effective reach is used in application of statistics to advertising and media analysis to calculate the effectiveness of ad and means used for ad. Effective reach is a time-dependent summary of aggregate audience behaviour.
Answer:fall arrest harness
Explanation:cuz it’s just right
Answer:
The electric field to balance the weight is approximately equals to 3.49x10^5 Newton/Coulumb
Explanation:
In order to be stationary position, magnitude of the total force due to electric field should be equal to the gravitational force that is, 
where

where <em>q</em> is the charge of the droplet and E is the electric field. On the other hand

where <em>m</em>,<em>V ,d</em> and<em> r</em> are the mass, volume, density and radius of the oil droplet respectively and <em>g </em>is the gravitational acceleration (g=9,80665 m/sn^2). By using the first equation and solving it for the electric field we can write,

(Note: d=0.85 x 10^-3 kg/cm^3 and the unit of electric field is Newton per coulumb (N/C))