can you please find the solution for que:4.10 Water10 mmWater depth1,80 m75 mm diameterValveTankbottom65 mmForce appliedto open valveValveHinge(a) General view of shower tank(b) Detail of valveand valveFIGURE 4.22 Shower tank and valve for Problem 4.10.49 The bottom of a laboratory vat has a hole in it to allow theliquid mercury to pour out. The hole is sealed by a rub-ber stopper pushed in the hole and held by friction. Whatforce tends to push the 0.75-in-diameter stopper out ofthe hole if the depth of the mercury is 28.0 in?4.10 A simple shower for remote locations is designed with acylindrical tank 500 mm in diameter and 1.800 m highas shown in Fig. 4.22. The water flows through a flappervalve in the bottom through a 75-mm-diameter opening.The flapper must be pushed upward to open the valve.How much force is required to open the valve?late the total force on the bottom of the closed tank3 mTank is1.2 m longAir 200 kPa (gage)Oil1.5m(sg = 0.80)Water26 m

Given data Diameter of cylinder = 500 mm = 0.5 m Height = 1.8 m Diameter of valve = 0.75 m Force…

410 A simple shower for remote locations is designed with a cylindrical tank 500 mm in diameter and 1.800 m high as shown in Fig. 4.22. The water flows through a flapper valve in the bottom through a 75-mm-diameter opening. The flapper must be pushed upward to open the valve. How much force is required to open the valve?

410 A simple shower for remote locations is designed with a cylindrical tank 500 mm in diameter and 1.800 m high as shown in Fig. 4.22. The water flows through a flapper valve in the bottom through a 75-mm-diameter opening. The flapper must be pushed upward to open the valve. How much force is required to open the valve?

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Similar questions

I need the answer as soon as possible
The Laws of Physics are written for a Lagrangian system, a well-defined system which we follow around – we will refer to this as a control system (CSys). For our engineering problems we are more interested in an Eulerian system where we have a fixed control volume, CV, (like a pipe or a room) and matter can flow into or out of the CV. We previously derived the material or substantial derivative which is the differential transformation for properties which are functions of x,y,z, t. We now introduce the Reynold’s Transport Theorem (RTT) which gives the transformation for a macroscopic finite size CV. At any instant in time the material inside a control volume can be identified as a control System and we could then follow this System as it leaves the control volume and flows along streamlines by a Lagrangian analysis. RTT:DBsys/Dt = ∂/∂t ʃCV (ρb dVol) + ʃCS ρbV•n dA; uses the RTT to apply the laws for conservation of mass, momentum (Newton's Law), and energy (1st Law of…
b) Figure Q4-b illustrates a gate of radius 4 m and of 1 direction of the resultant pressure force acting on the gate. Water surface S 2 m P Fig. Q4-b 30 4 m Support width. Find the magnitude and
Solved Examples 8 KN, Example 5 Replace the force and couple moment system acting on the beam in the figure below by an equivalent resultant force, and find where its line of action intersects the beam, measured from point O. 4 KN 15 KN m 0.5m Solution:- Force Summation. Summing the force components, 15 m 1.5 m + 1.5 m 1.5m (F), - XF: (F),- 8 KN() - 4.80 KNO
The magnitude of force F is 1000 N. Find the moment of force about: 1) Point B 2) Point O using: a) The Vector Method b) The Scalar Method *See attached image for problem 5 m B 3 m 2 m y 4 m 3 m