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julsineya [31]
4 years ago
15

The following displacement (cm), output (V) data have been recorded during a calibration of an LVDT. Displacement (cm), output (

V): 0.00 cm, 0.20 V; 0.125 cm, 0.60 V; 0.500 cm, 1.32 V; 0.769 cm, 1.95V; 1.019 cm; 2.33 V. Calculate the sensitivity of the LVDT expressed in (V/mm). (Please express your answer in V/mm and round up the answer to two (2) significant decimal digits.)

Engineering
1 answer:
nadya68 [22]4 years ago
4 0

Answer:

2.08V/cm

Explanation:

Plot the points on a graph. Draw the line of best fit. Calculate the gradietn of line of best fit.

Attached is the graph plotted on excel.

The equation of the line is

Votlage= 2.08× distance + 0.276

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Is it a problem that the stress intensity factor, K, from Irwin's near-tip approximation approaches infinity as you get close to
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8 0
1 year ago
A large class with 1,000 students took a quiz consisting of ten questions. To get an A, students needed to get 9 or 10 questions
VMariaS [17]

Answer:

a. 0.11

b. 110 students

c. 50 students

d. 0.46

e. 460 students

f. 540 students

g. 0.96

Explanation:

(See attachment below)

a. Probability that a student got an A

To get an A, the student needs to get 9 or 10 questions right.

That means we want P(X≥9);

P(X>9) = P(9)+P(10)

= 0.06+0.05=0.11

b. How many students got an A on the quiz

Total students = 1000

Probability of getting A = 0.11 ---- Calculated from (a)

Number of students = 0.11 * 1000

Number of students = 110 students

So,the number of students that got A is 110

c. How many students did not miss a single question

For a student not to miss a single question, then that student scores a total of 10 out of possible 10

P(10) = 0.05

Total Students = 1000

Number of Students = 0.05 * 1000

Number of Students = 50 students

We see that 5

d. Probability that a student pass the quiz

To pass, a student needed to get at least 6 questions right.

So we want P(X>=6);

P(X>=) =P(6)+P(7)+P(8)+P(9)+P(10)

=0.08+0.12+0.15+0.06+0.05=0.46

So, the probability of a student passing the quiz is 0.46

e. Number of students that pass the quiz

Total students = 1000

Probability of passing the quiz = 0.46 ----- Calculated from (d)

Number of students = 0.46 * 1000

Number of students = 460 students

So,the number of students that passed the test is 460

f. Number of students that failed the quiz

Total students = 1000

Total students that passed = 460 ----- Calculated from (e)

Number of students that failed = 1000 - 460

Number of students that failed = 540

So,the number of students that failed is 540

g. Probability that a student got at least one question right

This means that we want to solve for P(X>=1)

Using the complement rule,

P(X>=1) = 1 - P(X<1)

P(X>=1) = 1 - P(X=0)

P(X>=1) = 1 - 0.04

P(X>=1) = 0.96

7 0
3 years ago
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