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 × ( 1 / 100 ) ) m = 0.0150 m
density ρ = 790 kg/m³
Using expression for the gauged pressure;
p-p₀ = ρgh
where, p₀ is atmospheric pressure, ρ 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-p₀ = 790 kg/m³gh × 9.8 m/s² × 0.0150 m
= 116.13 ≈ 116 Pa
Therefore, the gauge pressure at the upper face of the block is 116 Pa
Answer:
-30m/s
Explanation:
Given:
Initial velocity of object = 200 feet/second
Final velocity of object = 50 feet/second
Time of travel = 5 seconds
To calculate acceleration of the object we will find the rate of change of velocity with respect to time.
So, acceleration "a" is given by:

where vf represents final velocity, vi represents initial velocity and is time of travel.
Plugging in values to evaluate acceleration.



The acceleration of the object is -30m/s