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3 years ago

10. True or False? A disruptive technology radically changes the way people live and work.

Engineering

2 answers:

Anettt [7]3 years ago

5 0

Answer:

True

Explanation:

According to Investopedia.com, "Disruptive technology is an innovation that significantly alters the way that consumers, industries, or businesses operate".

So yes, a disruptive does radically change the way people live and work.

Galina-37 [17]3 years ago

3 0

Answer: The technological changes that damage established companies are usually not radically new or difficult from a technological point of view. They do, however, have two important characteristics: First, they typically present a different package of performance attributes—ones that, at least at the outset, are not valued by existing customers. Second, the performance attributes that existing customers do value improve at such a rapid rate that the new technology can later invade those established markets. Only at this point will mainstream customers want the technology. Unfortunately for the established suppliers, by then it is often too late: the pioneers of the new technology dominate the market.

Explanation:

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) A brick of clay is being dried in a batch dryer using constant drying conditions. You have been asked to determine the amount

dimaraw [331]

Answer:

The drying time is calculated as shown

Explanation:

Data:

Let the moisture content be = 0.6

the free moisture content be = 0.08

total moisture of the clay = 0.64

total drying time for the period = 8 hrs

then if the final dry and wet masses are calculated, it follows that

t = (X0+ Xc)/Rc) + (Xc/Rc)* ln (Xc/X)

= 31.3 min.

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3 years ago

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natka813 [3]

Answer:

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Explanation:

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3 years ago

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scoundrel [369]

A is your answer choice

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3 years ago

Scissors is to cut as?

lidiya [134]

Answer:

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3 years ago

Consider a voltage v = Vdc + vac where Vdc = a constant and the average value of vac = 0. Apply the integral definition of RMS t

Anna11 [10]

Answer:

Proof is as follows

Proof:

Given that , $V = V_{ac} + V_{dc}$

for any function f with period T, RMS is given by

$RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[f(t)]^{2} } \, dt }$

In our case, function is $V = V_{ac} + V_{dc}$

$RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[V_{ac} + V_{dc}]^{2} } \, dt }$

Now open the square term as follows

$RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[V_{ac}^{2} + V_{dc}^{2} + 2V_{dc}V_{ac}] } \, dt }$

Rearranging terms

$RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {V_{dc}^{2} } \, dt + \frac{1}{T}\int\limits^T_0 {V_{ac}^{2} } \, dt + \frac{1}{T}\int\limits^T_0 {2V_{dc}V_{ac} } \, dt }$

You can see that

second term is square of RMS value of Vac

Third terms is average of VdcVac and given is that average of $V_{ac}V_{dc} = 0$

$RMS = \sqrt{\frac{1}{T}TV_{dc}^{2} + [RMS~~ of~~ V_{ac}]^2 }$

$RMS = \sqrt{V_{dc}^{2} + [RMS~~ of~~ V_{ac}]^2 }$

So it has been proved that given expression for root mean square (RMS) is valid

7 0

3 years ago