This law states that, despite chemical reactions or physical transformations, mass is conserved — that is, it cannot be created or destroyed — within an isolated system
Answer:
The water is flowing at the rate of 28.04 m/s.
Explanation:
Given;
Height of sea water, z₁ = 10.5 m
gauge pressure,
= 2.95 atm
Atmospheric pressure,
= 101325 Pa
To determine the speed of the water, apply Bernoulli's equation;
![P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2](https://tex.z-dn.net/?f=P_1%20%2B%20%5Crho%20gz_1%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Crho%20v_1%5E2%20%3D%20P_2%20%2B%20%5Crho%20gz_2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Crho%20v_2%5E2)
where;
P₁ = ![P_{gauge \ pressure} + P_{atm \ pressure}](https://tex.z-dn.net/?f=P_%7Bgauge%20%5C%20pressure%7D%20%2B%20P_%7Batm%20%5C%20pressure%7D)
P₂ = ![P_{atm}](https://tex.z-dn.net/?f=P_%7Batm%7D)
v₁ = 0
z₂ = 0
Substitute in these values and the Bernoulli's equation will reduce to;
![P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 = P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 = P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + \rho gz_1 = \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} + \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} + \rho gz_1)}{\rho} }](https://tex.z-dn.net/?f=P_1%20%2B%20%5Crho%20gz_1%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Crho%20v_1%5E2%20%3D%20%20P_2%20%2B%20%5Crho%20gz_2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Crho%20v_2%5E2%5C%5C%5C%5CP_1%20%2B%20%5Crho%20gz_1%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Crho%20%280%29%5E2%20%3D%20%20P_2%20%2B%20%5Crho%20g%280%29%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Crho%20v_2%5E2%5C%5C%5C%5CP_1%20%2B%20%5Crho%20gz_1%20%3D%20%20P_2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Crho%20v_2%5E2%5C%5C%5C%5CP_%7Bgauge%7D%20%2B%20P_%7Batm%7D%20%2B%20%5Crho%20gz_1%20%3D%20P_%7Batm%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Crho%20v_2%5E2%5C%5C%5C%5CP_%7Bgauge%7D%20%2B%20%20%5Crho%20gz_1%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%5Crho%20v_2%5E2%5C%5C%5C%5Cv_2%5E2%20%3D%20%5Cfrac%7B2%28P_%7Bgauge%7D%20%2B%20%20%5Crho%20gz_1%29%7D%7B%5Crho%7D%20%5C%5C%5C%5Cv_2%20%3D%20%5Csqrt%7B%20%5Cfrac%7B2%28P_%7Bgauge%7D%20%2B%20%20%5Crho%20gz_1%29%7D%7B%5Crho%7D%20%7D)
where;
is the density of seawater = 1030 kg/m³
![v_2 = \sqrt{ \frac{2(2.95*101325 \ + \ 1030*9.8*10.5 )}{1030} }\\\\v_2 = 28.04 \ m/s](https://tex.z-dn.net/?f=v_2%20%3D%20%5Csqrt%7B%20%5Cfrac%7B2%282.95%2A101325%20%5C%20%2B%20%5C%20%201030%2A9.8%2A10.5%20%29%7D%7B1030%7D%20%7D%5C%5C%5C%5Cv_2%20%3D%2028.04%20%5C%20m%2Fs)
Therefore, the water is flowing at the rate of 28.04 m/s.
Kinetic energy=1/2mv^2
=1/2(142*10^-3)(42.9)^2=130.6=131J
That’s really easy ask your teacher and also peace happy
Answer:
hope this helps i think the answer is C