Complete Question:
A horizontal tube consists of a 7.0-cm-diameter pipe, that narrows to a 2.0-cm-diameter throat. In the pipe, the water pressure is twice atmospheric pressure and the water flows with a speed of 0.40 m/s. What is the pressure in the throat, assuming that the water behaves like an ideal fluid. The density of water is 1000 kg/m^3, and the atmospheric pressure is 1.01 x 10^5 Pa.
Answer:
1.9 atm is the pressure in the throat
Explanation:
First find velocity of water and then find the pressure using Bernoulli's Principle.
Answer:
Explanation:
Work = Force times displacement. Therefore,
W = 3150(75.5) so
W = 238000 N*m
Answer:
In physics and engineering, a resultant force is the single force and associated torque obtained by combining a system of forces and torques acting on a rigid body. The defining feature of a resultant force, or resultant force-torque, is that it has the same effect on the rigid body as the original system of forces.
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Answer: Minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.
Explanation:
Velocity is only in horizontal direction at the top most point which is similar to the velocity in the horizontal direction at the time of launch.
Now, according to the law of conservation of energy the formula used is as follows.
As speed at which the person is travelling was 6.8 m/s. Hence, the initial velocity will be calculated as follows.
Thus, we can conclude that minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.
Answer:
The charge carried by the droplet is
Explanation:
Given that,
Distance =8.4 cm
Time = 0.250 s
Suppose tiny droplets of oil acquire a small negative charge while dropping through a vacuum in an experiment. An electric field of magnitude points straight down and if the mass of the droplet is
We need to calculate the acceleration
Using equation of motion
Put the value into the formula
We need to calculate the charge carried by the droplet
Using formula of electric filed
Put the value into the formula
Hence, The charge carried by the droplet is