Answer:
Eastward, at 11 m/s^2
Explanation:
64N-31N=unbalanced force of 33N
F=ma
33N=(3kg)a
a=11m/s^2 to the East
The air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625 N/m².
<h3 /><h3>What is pressure?</h3>
The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.
It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.
Pressure is found as the product of the density,acceleraton due to gravity and the height.
P₁=ρ₁gh₁
P₁=13,600 kg/m³×9.81 (m/s²)×0.18 m
P₁=24014.88 N/m²
P₂=ρ₂gh₂
P₂= 1000 kg/m³×9.81 (m/s²)×00.2 m
P₂=196.2 N/m²
P₃=ρ₃gh₃
P₃=850 kg/m³×9.81 (m/s²)×0.25
P₃=2084.625 N/m²
Hence,the air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625 N/m².
To learn more about the pressure refer to the link;
brainly.com/question/356585
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Answer:
p(a) * p(b) = .01923
p(b) = .01923 / .07692 = .2500
To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,
Through the aforementioned formula we will have to
The particulate part of the rest, so the final speed would be
Now from Newton's second law we know that
Here,
m = mass
a = acceleration, which can also be written as a function of velocity and time, then
Replacing we have that,
Therefore the force that the water exert on the man is 1386.62
