A hosepipe ejects water at a speed of 0.3m/s through a hole of area 50cm². If the water strikes the wall normally, calculate the force on the wall, assuming the velocity of the water normal to the wall is zero after striking.
Answer please with numerical calculations?
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F=M(V"-v')/t
F=V×d(0-v)/t
F=A×(L/t)×d×v
F=Av×v×d
F=A×V^2×d
here L=length
length/time=velocity
F=AV^2d
F=50×10^-4×0.3×0.3×1000
F=0.45N.
d=density of water.
A=area of pipe
v=velocity of water.
hope it is helpful.
F=V×d(0-v)/t
F=A×(L/t)×d×v
F=Av×v×d
F=A×V^2×d
here L=length
length/time=velocity
F=AV^2d
F=50×10^-4×0.3×0.3×1000
F=0.45N.
d=density of water.
A=area of pipe
v=velocity of water.
hope it is helpful.
I can't understand this please help
i meant area×square of velocity ×density
it is a formula as velocity after hitting is zero
i added the derivation of formula plz check it.
yeah now I understand
It was thankful
its ok
have a nice day
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F=50×10^-4×0.3×0.3×1000