FIGURE 0.13
umes
Observing total
water internal reflection in
ause water with a laser
age, beam refraction due
you to glans or beaker
hot neglected being very
thin).
Answers
Answered by
0
Answer:
Explanation:
True
Answered by
1
Explanation:
\begin{gathered}v_1=0.35\ m\ s^{-1}\quad\quad P_1=1\ cm\ Hg=1333\ Pa\\\\v_2=0.65\ m\ s^{-1}\quad\quad P_2=?\end{gathered}
v
1
=0.35 m s
−1
P
1
=1 cm Hg=1333 Pa
v
2
=0.65 m s
−1
P
2
=?
By Bernoulli's Theorem,
P_1+\dfrac {1}{2}\rho(v_1)^2=P_2+\dfrac {1}{2}\rho(v_2)^2P
1
+
2
1
ρ(v
1
)
2
=P
2
+
2
1
ρ(v
2
)
2
Potential energy is not considered since water flows through a horizontal pipe.
\begin{gathered}1333+\dfrac {1}{2}\times 1000(0.35)^2=P_2+\dfrac {1}{2}\times1000(0.65)^2\\\\\\\underline {\underline {P_2=1183\ Pa}}\end{gathered}
1333+
2
1
×1000(0.35)
2
=P
2
+
2
1
×1000(0.65)
2
P
2
=1183 Pa
Or,
\underline {\underline {P_2=0.887\ cm\ Hg}}
P
2
=0.887 cm Hg
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