Answer:
the gauge pressure at the upper face of the block is 116 Pa
Explanation:
Given the data in the question;
A cubical block of wood, 10.0 cm on a side.
height h = 1.50 cm = ( 1.50 Ć ( 1 / 100 ) ) m = 0.0150 m
density Ļ = 790 kg/m³
Using expression for the gauged pressure;
p-pā = Ļgh
where, pā is atmospheric pressure, Ļ is the density of the substance, g is acceleration due to gravity and h is the depth of the fluid.
we know that, acceleration due to gravity g = 9.8 m/s²
so we substitute
p-pā = 790 kg/m³gh Ć 9.8 m/s² Ć 0.0150 m
= 116.13 ā 116 Pa
Therefore, the gauge pressure at the upper face of the block is 116 Pa