the larger the cross sectional area, the smaller the resistance
The cross sectional area of a slab can be found by squaring the height of the slab.
reduction ratio= initial cross sectional area/final cross sectional area
Cross-sectional area = pi*radius2
To calculate the cross-sectional area of a shape, you need to determine the shape of the cross-section first (e.g., square, circle, triangle). Then, use the appropriate formula for that shape. For example, the formula for the cross-sectional area of a square is side length squared, for a circle it is pi times the radius squared, and for a triangle it is base times height divided by 2. Finally, plug in the given dimensions into the formula to calculate the cross-sectional area.
Three: 1) The area of the cross-sectional rectangle end 2) The area of the rectangle joining the longer side of the cross-sectional rectangular ends 3) The area of the rectangle joining the shorter side of the cross-sectional rectangular ends Then the surface area of the rectangular prism is twice the sum of these three areas.
the larger the cross sectional area, the smaller the resistance
Volume = cross sectional area * lengthArea = 2* cross sectional area + perimeter of cross section * length
The cross sectional area of a slab can be found by squaring the height of the slab.
Cross Sectional Area = Width x Average Depth
cross-sectional area = 0.5*(sum of parallel sides)*height
reduction ratio= initial cross sectional area/final cross sectional area
A Y12 bar typically has a cross-sectional area of 113 square millimeters.
cross sectional area of cable * voltage drop
The resistance of a wire is inversely proportional to the cross-sectional area of the wire. This means that as the cross-sectional area of the wire increases, the resistance decreases, and vice versa.
The answer depends on whether the cross sectional radius/diameter are doubles or the cross sectional area is doubled.
The answer depends on whether the cross sectional radius/diameter are doubles or the cross sectional area is doubled.