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2 years ago

10

list out main types of material used in design and Manufacture of product give one example for each in engineering application ?

1 answer:

cluponka [151]2 years ago

7 0

One of the types of product design is interface design. Another type of product design is called visual product design.

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You are given a noninverting 741 op-amp with a dc-gain of 23.6 dB. The input signal to this amplifier is;Vin(t) = (0.18)∙cos(2π(

Vsevolod [243]

Answer:

Output voltage equation is $V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)$

Explanation:

Given:

dc gain $A = 23.6$ dB

Input signal $V_{in} (t) = 0.18 \cos (2\pi (57000)t +18.3)$

Now convert gain,

$A = 10^{\frac{23.6}{20} } = 15.13$

DC gain at frequency $f = 0$ is given by,

$A = \frac{V_{out} }{V_{in} }$

$V_{out} =AV_{in}$

$V_{out} = 15.13 \times 0.18 \cos (2\pi (57000)t +18.3)$

At zero frequency above equation is written as,

$V_{out} = 2.72 \times \cos 18.3$

$V_{out} = 2.72$

Now we write output voltage as input voltage,

$V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)$

Therefore, output voltage equation is $V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)$

7 0

2 years ago

What can your employer do to protect you from overhead power lines?

agasfer [191]

Answer:

Have the power company install insulated sleeves (also known as “eels”) over power lines.

Wearing PPE is the only way to prevent being electrocuted

Explanation:

To prevent electrocution at workplace, employers can ensure that the power company install insulated sleeves (also known as “eels”) over power lines. Additionally, the employees should wear PPEs which are insulators to prevent electrocution.

5 0

3 years ago

The three resistors in the circuit shown have values of 4ohms 4ohms and 2ohms. The battery has a value of 12 volts. What is the

gtnhenbr [62]

Answer:

2ohms

Explanation:

3 0

3 years ago

The Energy Losses Associated with Valves and Fittings: a)- are generally associated with a K factor b)- are generally associated

madam [21]

Answer:

a)Are generally associated with factor.

Explanation:

We know that losses are two types

1.Major loss :Due to friction of pipe surface

2.Minor loss :Due to change in the direction of flow

As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and it produce losses in the energy.

Lot' of experimental data tell us that loss in the energy due to valve and fitting are generally associated with K factor.These losses are given as

$Losses=K\dfrac{V^2}{2g}$

8 0

3 years ago

Air enters the compressor of a simple gas turbine at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s. The compressor press

Zepler [3.9K]

Answer:

a) 3581.15067 kw

b) 95.4%

Explanation:

<u>Given data:</u>

compressor efficiency = 85%

compressor pressure ratio = 10

Air enters at: flow rate of 5m^3/s , pressure = 100kPa, temperature = 300 K

At turbine inlet : pressure = 950 kPa, temperature = 1400k

Turbine efficiency = 88% , exit pressure of turbine = 100 kPa

A) Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor in kW

attached below is a detailed solution to the given question

6 0

2 years ago