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.
Answer: ![115.52\ N](https://tex.z-dn.net/?f=115.52%5C%20N)
Explanation:
Given
Length of plank is 1.6 m
Force
is applied on the left side of plank
Force
is applied 43 cm from the left end O.
Mass of the plank is ![m=13.7\ kg](https://tex.z-dn.net/?f=m%3D13.7%5C%20kg)
for equilibrium
Net torque must be zero. Taking torque about left side of the plank
![\Rightarrow mg\times 0.8-F_2\times 0.43=0\\\\\Rightarrow F_2=\dfrac{13.7\times 9.8\times 0.8}{0.43}\\\\\Rightarrow F_2=249.78\ N](https://tex.z-dn.net/?f=%5CRightarrow%20mg%5Ctimes%200.8-F_2%5Ctimes%200.43%3D0%5C%5C%5C%5C%5CRightarrow%20F_2%3D%5Cdfrac%7B13.7%5Ctimes%209.8%5Ctimes%200.8%7D%7B0.43%7D%5C%5C%5C%5C%5CRightarrow%20F_2%3D249.78%5C%20N)
Net vertical force must be zero on the plank
![\Rightarrow F_1+W-F_2=0\\\Rightarrow F_1=F_2-W\\\Rightarrow F_1=249.78-13.7\times 9.8\\\Rightarrow F_1=115.52\ N](https://tex.z-dn.net/?f=%5CRightarrow%20F_1%2BW-F_2%3D0%5C%5C%5CRightarrow%20F_1%3DF_2-W%5C%5C%5CRightarrow%20F_1%3D249.78-13.7%5Ctimes%209.8%5C%5C%5CRightarrow%20F_1%3D115.52%5C%20N)
Answer:
I think the awnser is B (but don't qoute me on that) if its right then yay but if its wrong im sorry
Explanation:
The voltage from one side of the battery all the way around to the other side of the battery is 12v .
If 4 of those volts show up across the circle-thing, then the rest of the 12v ... 8v ... Must show up across the set of parallel rectangles.
To get that answer, I subtracted the 4 from the 12.
Just like it says in choice-C.