Answer:
When the imposter is sus : O
Explanation:
Answer:
Explanation:
You can utilize barbed clusters to store inadequate grids. On the off chance that there are a great many lines yet each line has just 4 or 5 associations with different segments, at that point as opposed to utilizing a 1000x1000 cluster you can utilize a 1000 line rough exhibit while you simply store the components that the present section has association with another segment. Other utilization can be done on account of query tables. Query tables will be tables which have different qualities concerning a solitary key where the quantity of qualities isn't fixed. Aside from this, barbed clusters have an exceptionally set number of utilization cases. Multidimensional exhibits then again have plenty of utilizations. It is utilized to store a great deal of information reliably on the grounds that the greater part of the information is put away is steady concerning which section compares to what information. Aside from that it very well may be utilized to make thick diagrams or sparse(not effective), plotting information. Another utilization case would be used as an impermanent stockpiling for the figurings that need to tail them and utilize the past information like in powerful programming.
Answer:
Explanation:
In this problem you need to define the force that acts upon a beam in a 3 point bending problem. I put a picture of the problem taken from Wikipedia:
In this problem the flexural strength is defined with the following formula:
where F is the force applied, L the length between the two rods, b the width of the ceramic block and d it's height.
The force is then defined as:
Answer:
The specific weight of unknown liquid is found to be 15 KN/m³
Explanation:
The total pressure in tank is measured to be 65 KPa in the tank. But, the total pressure will be equal to the sum of pressures due to both oil and unknown liquid.
Total Pressure = Pressure of oil + Pressure of unknown liquid
65 KPa = (Specific Weight of oil)(depth of oil) + (Specific Weight of unknown liquid)(depth of unknown liquid)
65 KN/m² = (8.5 KN/m³)(5 m) + (Specific Weight of Unknown Liquid)(1.5 m)
(Specific Weight of Unknown Liquid)(1.5 m) = 65 KN/m² - 42.5 KN/m²
(Specific Weight of Unknown Liquid) = (22.5 KN/m²)/1.5 m
<u>Specific Weight of Unknown Liquid = 15 KN/m³</u>
