Answer:
The velocity is 18.68m/s
Explanation:
Bernoulli's equation is applicable for stream line flow of a fluid. The flow must be steady and uniform flow. The Bernoulli's equation between inlet and outlet is written as:
P1/pg + V1/2g + Z1 = P2/pg + V2^2 + Z2
Where V1 and V2 are velocity of fluid at point 1 and 2b. The diameter of the tank too will be larger than that of the nozzle. Hence the velocity at point 1 will be 0.V1= 0
Substituting the values in to the equation
250 ×10^3/1000g + 0/g + 2.5 = 100×10^3/1000g + V2^2/2g + 0
250 + 2.5g = 100 + V2^2/2
250 + (2.5 × 9.8) = 100 V2^2/2
250 + 23.525- 100 = V2^2/2
174.525 = V2^2/2
Cross multiply
174.525 × 2 = V2^2
V2 = 349.05
V2 = Sqrt(349.05)
V2 = 18.68m/s
Answer:
d . ................. .........
It is given that u=36km/h
V= 72km/h
We have to convert it into m/s ao we will multiply it with 5/18
U=36×5/18=10
V=72×5/18=20
So acceleration =v-u/t
= 20-10/t
The time will be 9:45-9:25=20min
So we will convert it into second
= 20×60
a=20-10/20×60
=10/20×60
1/120
From newtons 2nd law
V^2-u^2=2as
S=v^2-u^2/2a
=400-100×120/2
=18000m/s
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