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alexandr1967 [171]
3 years ago
13

Suppose the measured background level is 5.1 mV. A signal of 20.7 mV is measured at a distance of 29 mm and 15.8 mV is measured

at 32.5 mm. Correct the data for background and normalize the data to the maximum value. What is the normalized corrected value at 32.5 mm
Engineering
1 answer:
Paladinen [302]3 years ago
3 0

Answer:  The normalized corrected value at 32.5 mm = 0.69 mV

Explanation:

Signal value V1 = 20.7 mV at

distance value = 29 mm and

V2 = 15.8 mV is measured at 32.5 mm. 

It is necessary, therefore, to subtract off this background level from the data to obtain a valid measurement

V1 = 20.7 - 5.1 = 15.6 mV

V2 = 15.8 - 5.1 = 10.7mV

The normalized corrected value at 32.5 mm will be

Vn = V2/V1 since V1 is the maximum value

Vn = 10.7/15.6 = 0.69 mV

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Harlamova29_29 [7]

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i

Explanation:

4 0
2 years ago
Let A→=(150iˆ+270jˆ) mm , B→=(300iˆ−450jˆ) mm , and C→=(−100iˆ−250jˆ) mm . Find scalars r and s, if possible, such that R→=rA→+s
ioda

Answer: r = 0.8081; s = -0.07071

Explanation:

A = (150i + 270j) mm

B = (300i - 450j) mm

C = (-100i - 250j) mm

R = rA + sB + C = 0i + 0j

R = r(150i + 270j) + s(300i - 450j) + (-100i - 250j) = 0i + 0j

R = (150r + 300s - 100)i + (270r - 450s - 250)j = 0i + 0j

Equating the i and j components;

150r + 300s - 100 = 0

270r - 450s - 250 = 0

150r + 300s = 100

270r - 450s = 250

solving simultaneously,

r = 0.8081 and s = -0.07071

QED!

5 0
3 years ago
What are the two reasons for a clear cut
Inessa [10]

Answer:

to clear land for agriculture and settlement and to use or sell timber for lumber, paper products, or fuel.

3 0
2 years ago
A 200-mm-long strip of metal is stretched in two steps, first to 300 mm and then to 400 mm. Show that the total true strain is t
Neko [114]

Explanation:

For true Strain:

step 1:

E true = Ln(1 + 0.5 ) = 0.40

Step 2:

E true = Ln(1 + 0.33 ) = 0.29

By single step process:

E true = Ln(1 + 1 ) = 0.69

total strain of step process = 0.40 + 0.29 = 0.69 units

SO TRUE STRAIN IS ADDITIVE.

4 0
2 years ago
Water flows through a nozzle at the end of a fire hose. If the nozzle exit velocity must be 20 m/s and the exit diameter is 40 m
lara31 [8.8K]

Answer:

minimum flow rate provided by pump is 0.02513 m^3/s

Explanation:

Given data:

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We know that flow rate Q is given as

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where A is Area

A =\frac{\pi}{4} \times (40\times 10^{-3})^2 = 1.256\times 10^{-3} m^2

Q = 1.256\times 10^{-3} \times 20 = 0.02513 m^3/s

minimum flow rate provided by pump is 0.02513 m^3/s

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3 years ago
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