The mass (in kg) of water in the pool is 1.264 
kg.
<h3>What is mass?</h3>
A numerical measurement of inertia, a basic characteristic of all matter, in physics. It essentially refers to a body of matter's resistance to changing its speed or location in response to the application of a force. The change caused by an applied force is smaller the more mass a body has.
Mass is a physical body's total amount of matter. Inertia, or the body's resistance to acceleration when a net force is applied, is also measured by this term. The strength of an object's gravitational pull to other bodies is also influenced by its mass. The kilogram serves as the SI's fundamental mass unit.
Given: A pool is 31.4 m long and 40.0 m wide. if the average depth of water is 3.30 ft
31.4 40.0 = 1256 
1 m = 39.37 in = 3.28 ft
3.30 ft = 1.01 m
Volume of pool = 1256 1.01 = 1264 
Since 1 water contains 
the volume of the pool is
1.264 E3 E3 = 1.264E6 kg
(or 1.264 kg)
The mass (in kg) of water in the pool is 1.264 kg.
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More expensive & probably less complicated
The final kinetic energy of the skateboarder after she freewheels and did work against friction on the flat section of the path is 4,600 J.
<h3>
Conservation of energy</h3>
The final kinetic energy of the stakeboarder is determined by applying the principle of conservation of energy as shown below;
ΔK.E = -W
K.Ef - K.Ei = -W
where;
- K.Ef is the final kinetic energy
- K.Ei is the initial kinetic energy
- W is work done
K.Ef = K.Ei - W
K.Ef = 5,000 J - 400 J
K.Ef = 4,600 J
Thus, the final kinetic energy of the skateboarder is 4,600 J.
Learn more about kinetic energy here: brainly.com/question/25959744
Answer:
t = (ti)ln(Ai/At)/ln(2)
t = 14ln(16)/ln(2)
Solving for t
t = 14×4 = 56 seconds
Explanation:
Let Ai represent the initial amount and At represent the final amount of beryllium-11 remaining after time t
At = Ai/2^n ..... 1
Where n is the number of half-life that have passed.
n = t/half-life
Half life = 14
n = t/14
At = Ai/2^(t/14)
From equation 1.
2^n = Ai/At
Taking the natural logarithm of both sides;
nln(2) = ln(Ai/At)
n = ln(Ai/At)/ln(2)
Since n = t/14
t/14 = ln(Ai/At)/ln(2)
t = 14ln(Ai/At)/ln(2)
Ai = 800
At = 50
t = 14ln(800/50)/ln(2)
t = 14ln(16)/ln(2)
Solving for t
t = 14×4 = 56 seconds
Let half life = ti
t = (ti)ln(Ai/At)/ln(2)
