Answer:
488.6KN
Explanation:
Hello!
the first step to solve this problem we must find the pressure exerted at the bottom of the tank (P) which is the sum of the external air pressure (P1 = 92kPa), the pressure inside the tank (P2 = 100kPa) and the pressure due to the weight of the water (P3), taking into account the above we have the following equation
P=P1+P2+P3
to find the pressure at the bottom of the tank due to the weight of the water we use the following equation

where
α=density=1 g/cm^3=1000kg/M^3
H=height=14.1m
g=gravity=3.71m/s^2
solving
P3=(1000)(14.1)(3.71)=52311Pa=52.3kPa
P=P1+P2+P3
P=100kPa+92kPa+52.3kPa=244.3kPa
finally to solve the problem we remember that the pressure is the force exerted on the area

Answer:
The angle of projection is 12.26⁰.
Explanation:
Given;
initial position of the dart, h₀ = 1.50 m
height above the ground reached by the dart, h₁ = 1.73 m
maximum height reached by the dart, Hm = h₁ - h₀ = 1.73 m - 1.50 m= 0.23 m
velocity of the dart, u = 10 m/s
The maximum height reached by the projectile is calculated as;

where;
θ is angle of projection
g is acceleration due to gravity = 9.8 m/s²

Therefore, the angle of projection is 12.26⁰.
The idea that <span>Max Planck propose to help explain why a blackbody radiator did not give off light of increasingly high frequency as its temperature increased is that </span>C. Matter can absorb light only in certain specific amounts.