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4 years ago

7

Legal metrology would protect consumers from businesses that do not take measurements according to defined measuring regulations

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1 answer:

sergij07 [2.7K]4 years ago

5 0

From what my research concluded, true

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Linear Time Invariant Systems For each of the systems below an input x(t) and the output y(t) are plotted. Determine whether eac

STALIN [3.7K]

Answer:

The system can be described by a convolution

Explanation:

Thinking process:

If we consider a discrete input to a linear time-invariant system, then the system will be periodic with respect to the period, say N. This therefore, means that the output must also be periodic. The proof is as follows:

The LTI system can be written for the system where:

y (n+N) = ∑ $h(k)x(n + N - k)$

= ∑ $h(k)x(n-k)\\= y(n)$

From the proof, it turns out that y(y + N) = y(n) for any value of n, then the output will be the periodic with the period N.

4 0

3 years ago

What engine does the Lexus is300 have?

Nikitich [7]

Answer:

2jz - ge

Explanation:

it has the non turbo 2jz. its the engine the supra has

6 0

4 years ago

Given V1 = 8 vpp, R1 = 6 kΩ, V2 = 5 vpp, R2 = 3 kΩ, V3 = 10 vpp, R3 = 3 kΩ, and Rf = 14 kΩ, find Vout. Vout = vpp. (Round your a

iragen [17]

Answer:

Vout= 93.3V

Explanation:

For this question, consider circuit in the attachment 1.

This is the circuit of an inverting amplifier. In an inverting amplifier

Vout/Vin= -Rf/Rin

To calculate the Vout, we must find Rin and Vin. For this we must solve the input circuit (attachment 2) using Thevinine theorem. Thevnine theorem states that all voltage sources in a circuit can be replaced by an equivalent voltage source Veq and and all resistances can be replaced by an equivalent resistance Req. To find out Req all voltage sources must be short circuited (attachment 3)

1/Req= 1/R1+1/R2+1/R3

1/Req=1/6+1/3+1/3

Req=6/5

To find out Veq consider circuit in attachment 4. We will solve this circuit using nodal analysis. In nodal analysis, we use the concept that sum of currents entering a node is equal to the sum of currents leaving a node. So,

I1= I2+I3

(10-Veq)/6= (Veq-5)/3+(Veq-10)/3

Veq=8V

Now the input circuit can be simplified as shown in attachment 5. Solve for Vout using equation

Vout/Veq= -Rf/Req

Vout/8= -14/(6/5)

Vout= - 93.3

It is at an angle of 180° from Veq

7 0

3 years ago

Technician A states that air tools generally produce more noise than electric tools, so wear ear protection when using air tools

dusya [7]

Answer:

Both Technician A and Technician B are correct

Explanation:

Air tools and electric tools are both power tools as they are used to make work easier. Air tools generally use an air compressor that powers the motor of the tool making it possible to use it while electric tools as the name implies are powered by an electric source which in this case is batteries. An example of an air tool is the nail gun which can be used by furniture makers to drive nails and they are often louder than electric tools because of vibrations caused by the compressor making it necessary to use ear protection when using the tool for ear safety.

Technician B is also correct because it is always advisable to use impact sockets while using impact guns due to the ability of the impact sockets to withstand the force caused by operating impact guns and make work neater when nuts and bolts are being loosened or tightened.

5 0

3 years ago

The steady-state data listed below are claimed for a power cycle operating between hot and cold reservoirs at 1200K and 400K, re

Anni [7]

Answer:

a) W_cycle = 200 KW , n_th = 33.33 % , Irreversible

b) W_cycle = 600 KW , n_th = 100 % , Impossible

c) W_cycle = 400 KW , n_th = 66.67 % , Reversible

Explanation:

Given:

- The temperatures for hot and cold reservoirs are as follows:

TL = 400 K

TH = 1200 K

Find:

For each case W_cycle , n_th ( Thermal Efficiency ) :

(a) QH = 600 kW, QC = 400 kW

(b) QH = 600 kW, QC = 0 kW

(c) QH = 600 kW, QC = 200kW

- Determine whether the cycle operates reversibly, operates irreversibly, or is impossible.

Solution:

- The work done by the cycle is given by first law of thermodynamics:

W_cycle = QH - QC

- For categorization of cycle is given by second law of thermodynamics which states that:

n_th < n_max ...... irreversible

n_th = n_max ...... reversible

n_th > n_max ...... impossible

- Where n_max is the maximum efficiency that could be achieved by a cycle with Hot and cold reservoirs as follows:

n_max = 1 - TL / TH = 1 - 400/1200 = 66.67 %

And, n_th = W_cycle / QH

a) QH = 600 kW, QC = 400 kW

- The work done by cycle according to First Law is:

W_cycle = 600 - 400 = 200 KW

- The thermal efficiency of the cycle is given by n_th:

n_th = W_cycle / QH

n_th = 200 / 600 = 33.33 %

- The type of process according to second Law of thermodynamics:

n_th = 33.333 % n_max = 66.67 %

n_th < n_max

Hence, Irreversible Process

b) QH = 600 kW, QC = 0 kW

- The work done by cycle according to First Law is:

W_cycle = 600 - 0 = 600 KW

- The thermal efficiency of the cycle is given by n_th:

n_th = W_cycle / QH

n_th = 600 / 600 = 100 %

- The type of process according to second Law of thermodynamics:

n_th = 100 % n_max = 66.67 %

n_th > n_max

Hence, Impossible Process

c) QH = 600 kW, QC = 200 kW

- The work done by cycle according to First Law is:

W_cycle = 600 - 200 = 400 KW

- The thermal efficiency of the cycle is given by n_th:

n_th = W_cycle / QH

n_th = 400 / 600 = 66.67 %

- The type of process according to second Law of thermodynamics:

n_th = 66.67 % n_max = 66.67 %

n_th = n_max

Hence, Reversible Process

7 0

3 years ago