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2 years ago

Which of the following is true about modern hydraulic lifts?

Engineering

1 answer:

kaheart [24]2 years ago

6 0

The modern hydraulic lifts make use of biodegradable fluid to transmit hydraulic power

Question: The options are left out in the question. The details and facts about the modern hydraulic lift are presented here

Details about the modern hydraulic lifts include;

The development of the modern hydraulic occurred in the Industrial Revolution to perform task done previously by steam powered elevators

The power of the hydraulic lift come from the hydraulic cylinder known as the actuator, which in turn is powered by pressurized hydraulic fluid such as oil

The hydraulic fluid is pushed by a piston rod through which energy is capable of being transferred, such that the applied force is multiplied, to provide more power for lifting

Facts about the modern hydraulic lifts include;

The dry motor in the modern hydraulic lift is more efficient and consumes 20% less energy

It comprises of valves that are controlled electronically such that the response is much rapid and the energy consumption is reduced by a further 20%

The cars used in the modern lift are lighter, as well as the slings, which reduces the power usage by 20%

It makes use of chemicals which are environmentally friendly as hydraulic fluid

The flash point of the fluid used is higher, as well as it posses 50% lower compressibility as well elasticity

Learn more here:

brainly.com/question/16942803

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3 years ago

When a man returns to his well-sealed house on a summer day, he finds that the house is at 35°C. He turns on the air conditioner

ipn [44]

Answer:

$\dot W = 1.667\, kW$

Explanation:

A well-sealed house means that there is no mass interaction between air indoors and outdoors. Hence, cooling process is isochoric. The heat removed by the air conditioner is:

$\dot Q_{L} = \frac{m_{air}}{\Delta t} \cdot c_{v, air} \cdot (T_{o}-T_{f})\\\dot Q_{L} = \frac{800\, kg}{(30\, min)\cdot (\frac{60\, s}{1 \, min} )}\cdot (0.7 \frac{kJ}{kg \cdot K}) \cdot (15\, K)\\\dot Q_{L} = 4.667\, kW$

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3 years ago

Suppose that we have a 1000pF parallel-plate capacitor with air dielectric charged to 1000V. The capacitor terminals are open ci

N76 [4]

Answer:

W = 1 mJ

V_new = 2000 V

W_new = 2 mJ

The extra energy came from the work done from moving the plates

Explanation:

We are given;

Capacitance; C = 1000pF = 10^(-9) F

Voltage; V = 1000V

Now,formula for stored energy in a parallel plate capacitor is given by;

W = (1/2)CV² = (1/2)(10^(-9))(1000²) = 0.5 mJ

However, in this case, it's W = 0.5 x 2 = 1 mJ since parallel-plate capacitor with air dielectric

When plates are moved and distance between plates is doubled(net charge is the same), thus we can calculate voltage from;

Q = CV

Since, C_new = C/2

Thus,

Q = (C/2)V_new

V_new = 2Q/C

Thus, V_new = 2V

Thus, V_new = 2 x 1000 = 2000 V

Now,

W_new = (1/2)C_new•(V_new)² = (1/2) (0.5C)•(2V)² = CV² = 2W = 2 x 1 = 2mJ

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3 years ago