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A jet of water squirts out horizontally from a hole near the bottom of the tank. The water shoots out .546 m from the tank and the water tank is held on a pedastol that is 1.28 m above the ground. What is the speed of the water coming out of the hole?


The second part is The hole has a diameter of 3.63 mm. What is the height of the water level in the tank?

2007-11-15 07:50:09 · 1 answers · asked by jULIAN C 2 in Science & Mathematics Physics

Alright, I get the first part, but what about the second?

2007-11-15 08:05:15 · update #1

1 answers

It is not that fast.

a )V=S/t
t=sqrt(2h/g) (the body will travel the same time in the vertical and horizontal directions)
V=S/sqrt(2h/g)
V=S sqrt(g/(2h))
V=.546 sqrt( 9.81/(2 x 1.28))
V=1.07 m/s

b) Potential energy at the top equal to kinetic energy at the botom
Pe=Ke
Pe= mgh ; Ke=0.5 mV^2
mgh =0.5 mV^2
gh =0.5 V^2
h= V^2/ (2g)
h=(1.07 )^2/(2 g)=
h=0.058m or 5.8 cm

2007-11-15 07:58:21 · answer #1 · answered by Edward 7 · 1 0

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