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

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For any object thrown upwards where only the force of gravity is acting upon it, uses the following formula for the maximum height attained.
H= v²/2g, where g = 9.81 m/s²
There are two information of velocities are given. However, we use the 20 m/s information because this is the launch velocity. Hence, the solution is as follows:
H = (20 m/s)²/2(9.81 m/s²)
<em>H = 20.4 m</em>
Answer
Pressure, P = 1 atm
air density, ρ = 1.3 kg/m³
a) height of the atmosphere when the density is constant
Pressure at sea level = 1 atm = 101300 Pa
we know
P = ρ g h


h = 7951.33 m
height of the atmosphere will be equal to 7951.33 m
b) when air density decreased linearly to zero.
at x = 0 air density = 0
at x= h ρ_l = ρ_sl
assuming density is zero at x - distance

now, Pressure at depth x


integrating both side


now,


h = 15902.67 m
height of the atmosphere is equal to 15902.67 m.
A force can be considered a push or pull
hope this helps :)
V=d/t
V=?
d=400m(4)
=1600m
t=6 min.
=360 s
V=1600m/360s
V=4.4m/s