A jet of water with cross section of 6 cm^2 s | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$F=\frac{d P}{d t}$

$=\frac{2 d m V \cos 60^{\circ}}{d t}$

$=\frac{2(\rho A d x) V \cos 60^{\circ}}{d t}$

$=2 \rho A V^{2} \cos 60^{\circ}$

Vedclass · Accepted Answer

$86.4 $

Vedclass · Answer

$F=\frac{d P}{d t}$

$=\frac{2 d m V \cos 60^{\circ}}{d t}$

$=\frac{2(ho A d x) V \cos 60^{\circ}}{d t}$

$=2 ho A V^{2} \cos 60^{\circ}$

$=10^{3} 	imes 6 	imes 10^{-4} 	imes 12^{2}$

$=86.4 \;N$

A jet of water with cross section of $6$ $cm^2$ strikes a wall at an angle of $60^o $ to the normal and rebounds elastically from the wall without losing energy. If the velocity of the water in the jet is $12$ $m/s$, the force acting on the wall is ....... $N$

Similar Questions

The density of the atmosphere is $1.29\, kg/m^3$, then how high would the atmosphere extend ? $(g = 9.81\, m/sec^2)$ ........ $km$

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities $d_1$ and $d_2$ are filled in the tube. Each liquid subtends $90^o$ angle at centre. Radius joining their interface makes an angle $\alpha$ with vertical. Ratio $\frac{{{d_1}}}{{{d_2}}}$ is

Mention two instruments which measure the atmospheric pressure.

Who devised for first time, a method for measuring atmospheric pressure ?

Radius of an air bubble at the bottom of the lake is $r$ and it becomes $ 2r $ when the air bubbles rises to the top surface of the lake. If $P $ $cm$ of water be the atmospheric pressure, then the depth of the lake is