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2 years ago

Which of the following can easily reverse motion and are better at varying speeds than electrical motors?

Engineering

2 answers:

yan [13]2 years ago

6 0

Answer:

fluid power system

Explanation:

satela [25.4K]2 years ago

4 0

Fluid power system

Example

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Which of the following is NOT true about hydraulic systems?

Dmitry [639]

Answer: The answer is D

D.In hydraulic systems, the operating temperatures must be kept between 170�F and 180°F

Explanation:

The operating temperature for hydraulic systems is 140°F and below. Anything above this temperature is too high and will reduce the useful life of hydraulic fluid.

Most often problems associated with hydraulic systems are caused by fluid contaminated with particulate matter.

7 0

3 years ago

A car travells at 67.5 km\h in 120 km.how long will it take to reach the destination

Kruka [31]

mark me the brainiest here

average speed (in km/h) of a car stuck in traffic that drives 12 kilometers in 2 hours.

5 0

3 years ago

Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW o

jekas [21]

Complete Question

Complete Question is attached below.

Answer:

$V'=5m/s$

Explanation:

From the question we are told that:

Diameter $d=0.10m$

Power $P=4.0kW$

Head loss $\mu=10m$

$\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu$

$\frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10$

$H_m=(\frac{200*10^3}{1000*9.8}-10)$

$H_m=10.39m$

Generally the equation for Power is mathematically given by

$P=\rho gQH_m$

Therefore

$Q=\frac{P}{\rho g H_m}$

$Q=\frac{4*10^4}{1000*9.81*10.9}$

$Q=0.03935m^3/sec$

Since

$Q=AV'$

Where

$A=\pi r^2\\A=3.142 (0.05)^2$

$A=7.85*10^{-3}$

Therefore

$V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}$

$V'=5m/s$

5 0

3 years ago

The hot combustion gases of a furnace are separated from the ambient air and its surroundings, which are at 25 oC, by a brick wa

yanalaym [24]

Answer:

T1 = 625.54 K

Explanation:

We are given;

T_α = Tsur = 25°C = 298K

h = 20 W/m².K,

L = 0.15 m

K = 1.2 W/m.K

ε = 0.8

Ts = T2 = 100°C = 373K

T1 = ?

Assumption:

-Steady- state condition

-One- dimensional conduction

-No uniform heat generation

-Constant properties

From Energy balance equation;

E°in - E°out = 0

Thus,

q"cond – q"conv – q"rad = 0

K[(T1 - T2)/L] - h(Ts-T_α) - εσ (Ts⁴ – Tsur⁴)

Where σ is Stephan-Boltzmann constant and has a value of 5.67 x 10^(-8)

Thus;

K[(T1 - T2)/L] - h(Ts-T_α) - εσ (Ts⁴ – Tsur⁴) = 1.2[(T1 - 373)/0.15] - 20(373 - 298] - 0.8x5.67x10^(-8)[373⁴ - 298⁴] = 0

This gives;

(8T1 - 2984) - (1500) - 520.31 = 0

8T1 = 2984 + 1500 + 520.31

8T1 = 5004.31

T1 = 5004.31/8

T1 = 625.54 K

7 0

3 years ago

Design a filter that has infinite DC gain, a gain of one from 1Hz to 100 Hz and filters (1storder) any signals above 100 Hz.a) S

EastWind [94]

Answer:

Attached below are the sketches

answer :

c) G(s) = 100 / ( s + 100 )

d) y'(t) + 100Y(s) = 100 X(s)

e) g(t) = e^-100t u(t)

Explanation:

a) Sketch the bode plot

The filter here is a low pass filter

b) Sketch the s-plane

attached below. pole ( s ) is at 100

c) write the transfer function of the filter

Transfer function ; G(s) = 100 / ( s + 100 )

d) write the differential equation

Y(s) / X(s) = 100 / s + 100

Y(s) [ s + 100 ] = 100 X(s)

= sY(s) + 100Y = 100 X(s)

∴ differential equation = y'(t) + 100Y(s) = 100 X(s)

e) write out the unforced transient response

g(t) = e^-100t u(t)

f) write out the frequency response

attached below

4 0

3 years ago