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valentina_108 [34]
3 years ago
15

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

Engineering
2 answers:
yan [13]3 years ago
6 0

Answer:

fluid power system

Explanation:

satela [25.4K]3 years ago
4 0
Fluid power system

Example
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An engineer lives in Hawaii at a location where the annual rain fall is 300 inches. She decides to use the rain to generate elec
Alex777 [14]

Answer:

80.7lbft/hr

Explanation:

Flow rate of water in the system = 3.6x10^-6

The height h = 100

1s = 1/3600h

This implies that

Q = 3.6x10^-6/[1/3600]

Q = 0.0000036/0.000278

Q = 0.01295

Then the power is given as

P = rQh

The specific weight of water = 62.3 lb/ft³

P = 62.3 x 0.01295 x 100

P = 80.675lbft/h

When approximated

P = 80.7 lbft/h

This is the average power that could be generated in a year.

This answers the question and also corresponds with the answer in the question.

4 0
3 years ago
Find a negative feedback controller with at least two tunable gains that (1) results in zero steady state error when the input i
IceJOKER [234]

Answer:

Gc(s) = \frac{0.1s + 0.28727}{s}

Explanation:

comparing the standard approximation with the plot attached we can tune the PI gains so that the desired response is obtained. this is because the time requirement of the setting is met while the %OS requirement is not achieved instead a 12% OS is seen from the plot.

attached is the detailed solution and the plot in Matlab

8 0
4 years ago
How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an
iris [78.8K]

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

thus,

Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

or

Q = 18.615 × 10⁻¹⁸ C

Also,

We know

Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

4 0
3 years ago
The 50mm diameter cylinder is made from Am 1004-T61 magnesium (E = 44.7GPa, a = 26x10^-6/°C)
BartSMP [9]

Answer:

......................

Explanation:

3 0
3 years ago
Problem 34.3 The elevation of the end of the steel beam supported by a concrete floor is adjusted by means of the steel wedges E
Natasha2012 [34]

The image is missing, so i have attached it.

Answer:

A) P = 65.11 KN

B) Q = 30 KN

Explanation:

We are given;

The end reaction of the beam; F = 100KN

Coefficient of static friction between two steel surfaces;μ_ss = 0.3

Coefficient of static friction between steel and concrete;μ_sc = 0.6

So, F1 = μ_ss•F =0.3 x 100 = 30 KN

F2 = μ_ss•N_EF = 0.3N_EF

From the screen shot, we see that the angle is 12°

Sum of forces in the Y-direction gives;

F2•sin12 - N_EF•cos12 + 100 = 0

Rearranging gives;

N_EF•cos12 - F2•sin12 = 100

Let's put 0.3N_EF for F2 to give;

N_EF•cos12 - 0.3N_EF•sin12 = 100

Thus;

N_EF(0.9158) - 0.1247 = 100

N_EF(0.9781) = 100 + 0.1247

N_EF = 100.1247/0.9158

N_EF = 109.33 KN

Thus, F2 = 0.3N_EF = 0.3 x 109.33 = 32.8 KN

Wedge will move if;

P = (F1 + F2cos12 + N_EFsin12)

Thus;

P = 10 + (32.8 x 0.9781) + (109.33 x 0.2079)

P ≥ 65.11 KN

B) For static equilibrium, Q = F1

Thus, Q = 30 KN

3 0
4 years ago
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