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Leviafan [203]
2 years ago
12

How much work, in Newtons, is required to lift a 20.4-kg (45lb) plate from the ground to a stand that is 1.50 meters up?

Engineering
1 answer:
nataly862011 [7]2 years ago
4 0

Answer:

Explanation:

Work, U, is equal to the force times the distance:

U = F · r

Force needed to lift the weight, is equal to the weight: F = W = m · g

so:

U = m · g · r

   = 20.4kg · 9.81 \frac{N}{kg} · 1.50m

   = 35.316 \frac{N}{m}

   = 35.316 W

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WHAT IS A TOROID IN HYDRAULUCS?
xenn [34]

Answer:

A hydraulic actuator comprised of a toroidal piston within a toroidal enclosure, with differential fluid pressure alternatively applied to upper and lower surfaces of the piston to cause a reciprocating motion, and with plural double ended piston rods extending in parallel above and below the piston, and slidably extending in fluid sealed relation through end caps of the toroidal enclosure to distribute the vibrational force produced by the reciprocating piston over plural points of a surface area of the mass to be vibrated, thereby reducing the likelihood of stress, strain, or harmonics in the mass.

3 0
3 years ago
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Which - type of service shop is least likely to provide service to all
MatroZZZ [7]
E. None of the above
8 0
2 years ago
The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is
romanna [79]

Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

a)

For cylinder AB:

Let Length of AB = 12 in

c=\frac{1}{2}d=\frac{1}{2} *1.1=0.55in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}0.55^4=0.1437\ in^4\\

\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437}  =2.53*10^{-5}T_{AB}

For cylinder BC:

Let Length of BC = 18 in

c=\frac{1}{2}d=\frac{1}{2} *2.2=1.1in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}1.1^4=2.2998\ in^4\\

\phi_B=\frac{T_{BC}L}{GJ}=\frac{T_{BC}*18}{5.9*10^6*2.2998}  =1.3266*10^{-6}T_{BC}

2.53*10^{-5}T_{AB}=1.3266*10^{-6}T_{BC}\\T_{BC}=19.0717T_{AB}

T_{AB}+T_{BC}-T=0\\T_{AB}+T_{BC}=T\\T_{AB}+T_{BC}=14*10^3\ lb.in\\but\ T_{BC}=19.0717T_{AB}\\T_{AB}+19.0717T_{AB}=14*10^3\\20.0717T_{AB}=14*10^3\\T_{AB}=0.697*10^3\ lb.in\\T_{BC}=13.302*10^3\ lb.in

b) Maximum shear stress in BC

\tau_{BC}=\frac{T_{BC}}{J}c=13.302*10^3*1.1/2.2998=6.352\ ksi

Maximum shear stress in AB

\tau_{AB}=\frac{T_{AB}}{J}c=0.697*10^3*0.55/0.1437=2.667\ ksi

8 0
3 years ago
Steam enters an adiabatic condenser (heat exchanger) at a mass flow rate of 5.55 kg/s where it condensed to saturated liquid wat
Evgen [1.6K]

Answer:

The minimum mass flow rate will be "330 kg/s".

Explanation:

Given:

For steam,

m_{s}=5.55 \ kg/s

\Delta h=2491 \ kg/kj

For water,

\Delta T=10^{\circ}C

(Cp)_{w}=4.184 \ kJ/kg^{\circ}C

They add energy efficiency as condenser becomes adiabatic, with total mass flow rate of minimal vapor,

⇒  m_{s}\times (\Delta h)=M_{w}\times(Cp)_{w}\times \Delta T

On putting the estimated values, we get

⇒  5.55\times 2491=M_{w}\times 4.184\times 10\\

⇒  13825.05=M_{w}\times 41.84

⇒  M_{w}=330 \ kg/s

7 0
3 years ago
. In one stroke of a reciprocating compressor, helium is isothermally and reversibly compressed in a piston + cylinder from 298
andriy [413]

Answer:

5.7058kj/mole

Explanation:

Please see attachment for step by step guide

5 0
3 years ago
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