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2 years ago

Which of the following fluid power systems is portable and human-scale?

Engineering

2 answers:

meriva2 years ago

7 0

Answer:

a fluid power engine okk done please

faltersainse [42]2 years ago

3 0

Answer:

A fluid powered drill

Explanation:

Firefighters use these at car accidents because of their durability reliability and incredible torque

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In the engineering design process, what do engineers do immediately after brainstorming multiple solutions?

34kurt

Answer:

Iterate to improve the solution

7 0

3 years ago

A piston–cylinder assembly contains 5 kg of air, initially at 2.0 bar, 30 C. The air undergoes a process to a state where the pr

Ainat [17]

Answer: wor done is 145. 06kJ

Heat transfer is 135.53kJ

Explanation:

No of moles of air = mass/molar mass = 5000g/28gmol^-1 = 172.65mol

P1 = 2bar =2*101300 =202600pa

T1 = 30° +273k = 303k

P2 =p1 = 202600pa

V2 =? T2 =?

Using pV = nRT

R = 8.314 PA m^3 mol^-1 k^-1

V1 = (172.65*8.314*303)/202600

V1 = 2.146m^3

For second state, 1.5pv = const = P1V1

V2 = (202600*2.146)/(1.5*202600)

V2 = 1.43m^3

Volume change = 2.146 - 1.43 =0.715m^3

Word done = pressure* volume change

W = 202600*0.716 = 145061.6J

= 145.061kJ

Using V1/T1 = V2/T2

T2 = V2T1/V1

=(1.43*303)/2.146 = 201.9k

For internal energy U

U = nCv*(T2 - T1)

*CV is the heat capacity at const. vol approximately 0.718J mol^-1 k^-1

U = 172.65*0.718*(201.9-303)

U = -12532.6J = -12.532kJ

The -ve means the system lost internal energy.

Q = U+W = total heat energy of system

Q = - 12.532+145.061 = 132.52 kJ

7 0

4 years ago

A large well-mixed tank of unknown volume, open to the atmosphere initially, contains pure water. The initial height of the solu

trasher [3.6K]

Answer:

The exact time when the sample was taken is = 0.4167337 hr

Explanation:

The diagram of a sketch of the tank is shown on the first uploaded image

Let A denote the first inlet

Let B denote the second inlet

Let C denote the single outflow from the tank

From the question we are given that the diameter of A is = 1 cm = 0.01 m

Area of A is = $\frac{\pi}{4}(0.01)^{2} m^{2}$

= $7.85 *10^{-5}m^{2}$

Velocity of liquid through A = 0.2 m/s

The rate at which the liquid would flow through the first inlet in terms of volume = $\frac{Volume of Inlet }{time}$ = $Velocity * Area$ i.e is $m^{2} * \frac{m}{s} = \frac{m^{3}}{s}$

= $0.2 *7.85*10^{-5} \frac{m^{3}}{s}$

The rate at which the liquid would flow through the first inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

= $1039.8 * 0.2 * 7.85 *10^{-5} Kg/s$

= 0.016324 $\frac{Kg}{s}$

From the question the diameter of B = 2 cm = 0.02 m

Area of B = $\frac{\pi}{4} * (0.02)^{2} m^{2} = 3.14 * 10^{-4}m^{2}$

Velocity of liquid through B = 0.01 m/s

The rate at which the liquid would flow through the first inlet in terms of volume = $\frac{Volume of Inlet }{time}$ = $Velocity * Area$ i.e is $m^{2} * \frac{m}{s} = \frac{m^{3}}{s}$

= $3.14*10^{-4} *0.01 \frac{m^{3}}{s}$

The rate at which the liquid would flow through the second inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

= $1053 * 3.14*10^{-6} \frac{Kg}{s}$

= 0.00330642 $\frac{Kg}{s}$

From the question The flow rate in term of volume of the outflow at the time of measurement is given as = 0.5 L/s

And also from the question the mass of potassium chloride at the time of measurement is given as 13 g/L

So The rate at which the liquid would flow through the outflow in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

= $13\frac{g}{L} * 0.5 \frac{L}{s}$

= $\frac{6.5}{1000}\frac{Kg}{s}$ Note (1 Kg = 1000 g)

= 0.0065 kg/s

Considering potassium chloride

Let denote the rate at which liquid flows in terms of mass as as $\frac{dm}{dt}$ i.e change in mass with respect to time hence

Input(in terms of mass flow ) - output(in terms of mass flow ) = Accumulation in the Tank(in terms of mass flow )

(0.016324 + 0.00330642) - 0.0065 = $\frac{dm}{dt}$

$\int\limits {\frac{dm}{dt} } \, dx =\int\limits {0.01313122} \, dx$

=> 0.01313122 t = (m - $m_{o}$ )

From the question (m - $m_{o}$ ) is given as = 19.7 Kg

Hence the time when the sample was taken is given as

0.01313122 t = 19.7 Kg

=> t = 1500.2414 sec

t = .4167337 hours (1 hour = 3600 seconds)

5 0

4 years ago

What is the condition for sampling frequency to reconstruct the information signal ?

bagirrra123 [75]

A continuous time-varying 1-D signal is sampled by narrow sampling pulses at a regular rate fr = 1/T, which must be at least twice the bandwidth of the signal. At first, it may be somewhat surprising that the original waveform can be reconstructed exactly from a set of discrete samples.

8 0

3 years ago

Instructions: For each problem, identify the appropriate test statistic to be use (t test or z-test). Then compute z or t value.

Natalka [10]

The sample of 81 students was selected with a mean score of 90, this illustrates an example of a right tailed one sample z test.

<h3>How to illustrate the sample?</h3>

From the information given, the teacher claims that the mean score of students in his class is greater than 82 with a standard deviation of 20. In this case, the right tailed one sample z test is used.

The z value will be:

= (90 - 82)/20/✓81

= 3.6

Since 3.6 > 1.645, the null hypothesis will be rejected as there's enough evidence to support the teacher's claim.

When an online medicine shop claims that the mean delivery time for medicines is less than 120 minutes with a standard deviation of 30 minutes, the left tailed one sample test is used.

The z value will be:

= (100 - 120)/(30/✓49)

= -4.66

The null hypothesis is rejected as there is enough evidence to support the claim of the medicine shop.

Learn more about sampling on:

brainly.com/question/17831271

#SPJ1

6 0

2 years ago