The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as
<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>
Generally, the equation for Rate of flow of Liquid is mathematically given as
$$
Where dP is pressure difference r is the radius
is the viscosity of water
L is the length of the pipe
In $30s the quantity that flows out of the tube
In conclusion, the quantity that flows out of the tube
Read more about the flows rate
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Freezing.................
Answer:
= 6.55cm
Explanation:
Given that,
distance = 1.26 m
distance between two fourth-order maxima = 53.6 cm
distance between central bright fringe and fourth order maxima
y = Y / 2
= 53.6cm / 2
= 26.8 cm
=0.268 m
tan θ = y / d
= 0.268 m / 1.26 m
= 0.2127
θ = 12°
4th maxima
d sinθ = 4λ
d / λ = 4 / sinθ
d / λ = 4 / sin 12°
d / λ = 19.239
for first (minimum)
d sinθ = λ / 2
sinθ = λ / 2d
= 1 / 2(19.239)
= 1 / 38.478
= 0.02599
θ = 1.489°
tan θ = y / d
y = d tan θ
= 1.26 tan 1.489°
= 0.03275
the total width of the central bright fringe
Y = 2y
= 2(0.03275)
= 0.0655m
= 6.55cm
Answer:
1 μC extra charge will be flow here
Explanation:
Given data
battery V1 = 4.0 V
flows Q1 = 6.0 μC
replace battery V2 = 7.0 V
to find out
what happen if we replace battery
solution
we apply here principal of capacitor
that is Q directly proportional voltage
so we say Q2/Q1 = V2/ V1
put all value here
Q2/Q1 = V2/ V1
Q2/6 = 7/ 6
Q2 = 7
so we see here 7 μC will be flow
and Q = Q2 - Q1 = 7 - 6 = 1 μC
so we also say that 1 μC extra charge will be flow here
