Thus pressure
${p}_{2}$ over the second opening is reduced by
$\frac{1}{2}\rho {v}_{2}^{2}$ , so the fluid in the manometer rises by
h on the side connected to the second opening, where
$h\propto \frac{1}{2}\rho {v}_{2}^{2}.$
(Recall that the symbol
$\propto $ means “proportional to.”) Solving for
${v}_{2}$ , we see that
${v}_{2}\propto \sqrt{h}.$
Part (b) shows a version of this device that is in common use for measuring various fluid velocities; such devices are frequently used as air-speed indicators in aircraft.
A fire hose
All preceding applications of Bernoulli’s equation involved simplifying conditions, such as constant height or constant pressure. The next example is a more general application of Bernoulli’s equation in which pressure, velocity, and height all change.
Calculating pressure: a fire hose nozzle
Fire hoses used in major structural fires have an inside diameter of 6.40 cm (
[link] ). Suppose such a hose carries a flow of 40.0 L/s, starting at a gauge pressure of
$1.62\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{6}{\text{N/m}}^{2}$ . The hose rises up 10.0 m along a ladder to a nozzle having an inside diameter of 3.00 cm. What is the pressure in the nozzle?
Strategy
We must use Bernoulli’s equation to solve for the pressure, since depth is not constant.
where subscripts 1 and 2 refer to the initial conditions at ground level and the final conditions inside the nozzle, respectively. We must first find the speeds
${v}_{1}$ and
${v}_{2}$ . Since
$Q={A}_{1}{v}_{1}$ , we get
This value is a gauge pressure, since the initial pressure was given as a gauge pressure. Thus, the nozzle pressure equals atmospheric pressure as it must, because the water exits into the atmosphere without changes in its conditions.
Bernoulli’s equation states that the sum on each side of the following equation is constant, or the same at any two points in an incompressible frictionless fluid:
Bernoulli’s principle is Bernoulli’s equation applied to situations in which the height of the fluid is constant. The terms involving depth (or height
h ) subtract out, yielding
bcoz it doesn't satisfy the algabric laws of vectors
Shiekh
The Electric current can be defined as the dot product of the current density and the differential cross-sectional area vector : ... So the electric current is a scalar quantity . Scalars are related to tensors by the fact that a scalar is a tensor of order or rank zero .
*a plane flies with a velocity of 1000km/hr in a direction North60degree east.find it effective velocity in the easterly and northerly direction.*
imam
hello
Lydia
hello Lydia.
Sackson
What is momentum
isijola
hello
Saadaq
A rail way truck of mass 2400kg is hung onto a stationary trunk on a level track and collides with it at 4.7m|s. After collision the two trunk move together with a common speed of 1.2m|s. Calculate the mass of the stationary trunk
principle of superposition allows us to find the electric field on a charge by finding the x and y components
Kidus
Two Masses,m and 2m,approach each along a path at right angles to each other .After collision,they stick together
and move off at 2m/s at angle 37° to the original direction of the mass m. What where the initial speeds of the two particles
MB
2m & m initial velocity 1.8m/s & 4.8m/s respectively,apply conservation of linear momentum in two perpendicular directions.
Shubhrant
A body on circular orbit makes an angular displacement given by teta(t)=2(t)+5(t)+5.if time t is in seconds calculate the angular velocity at t=2s
MB
2+5+0=7sec
differentiate above equation w.r.t
time, as angular velocity is rate of change of angular displacement.
Shubhrant
Ok i got a question I'm not asking how gravity works. I would like to know why gravity works. like why is gravity the way it is. What is the true nature of gravity?
gravity pulls towards a mass...like every object is pulled towards earth
Ashok
An automobile traveling with an initial velocity of 25m/s is accelerated to 35m/s in 6s,the wheel of the automobile is 80cm in diameter. find
* The angular acceleration