32) webassign.net/webistudent/Assignment-ResponsesMast?dep-1567557 igneous Rocks

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32) webassign.net/webistudent/Assignment-ResponsesMast?dep-1567557 igneous Rocks Less MACRoeconomics.. Macro Unit 3 summ syllabi for Electrical Apps Guide 5552 -Apply- What is this letter u- IMM 5558 E Docu.. D According to our records you have not yet redeemed an access code for this class or purchased access online. The grace period will end Wednesday, January 18 2017 at 1200 AM CST. After that date you wil no longer be able to see your WebAssign assignments or grades Getaccess now. Assignment scoring Your last submission is used for your score. My Notes Ask Your Teache -1 points SerPSE9 14 P circular have a radius of 0.400 A large man sits on a four-legged with the floor. The combined mass of the man and chair is 9 kg. If the chair legs are and chair his feet off cm at the bottom, what pressure does each leg exert on the fioor? My Notes o Ask your T 2. O -1 points sorpsEg 14 pooawi. The small piston of a hydraulic lift (see figure below) has area 2.7o cm2, and its large piston has a cross-sectional area of 201 cm What downward force of magnitude F1 must be applied to the small piston for a is Fo 15,2 kN? the in to raise a load whose weight

Explanation / Answer

Given Data

Combined mass, m = 95 kg

radius = 0.4 cm = 0.4*10^-2 m

Solution : -

Pressure = F/area

Here

Force = m * g

Area = *r2

and Weight exerted for each leg = combined mass / No. of Legs

                                                = 95/4

                                                = 23.75 kg

So Force = m*g

              = 23.75*9.8

              = 232.75

Area = *r2

        = *(0.4*10^-2 m)^2

        = 5.026 *10^-5 m^2

So Pressure = F/area

P = F/A

P = 232.75 / 5.026 *10^-5

= 4.63*10^6 Pa --> Answer

Q-2)

Given Data

Area of cross-section of small piston, A1 = 2.7 cm^2

Area of cross-section of large piston, A2 = 201 cm^2

Load Weight, F2 = 15.2 kN

Force on small piston, F1 = ?

Solution:-

From Pascals Law, we have

P 1 = P2

F1/A1 = F2/A2

F1 = (A1/A2)*F2

     = (2.7 / 201) * 15.2 kN

     = 0.204 kN = 204.2 N

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