All categories

3 years ago

SRSS is a method used in responce spectram analysis

Engineering

1 answer:

VMariaS [17]3 years ago

5 0

Answer:

Modal analysis is performed to identify the modes, and the response in that mode can be picked from the response spectrum. These peak responses are then combined to estimate a total response. A typical combination method is the square root of the sum of the squares (SRSS) if the modal frequencies are not close.

You might be interested in

The Environmental Protection Agency (EPA) has standards and regulations that says that the lead level in soil cannot exceed the

DENIUS [597]

Answer:

See below

Explanation:

Check One-Sample T-Interval Conditions

Random Sample? √

Sample Size ≥30? √

Independent? √

Population Standard Deviation Unknown? √

One-Sample T-Interval Information

Formula --> $CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})$
Sample Mean --> $\bar{x}=390.25$
Critical Value --> $t^*=2.0096$ (given $df=n-1=50-1=49$ degrees of freedom at a 95% confidence level)
Sample Size --> $n=50$
Sample Standard Deviation --> $S_x=30.5$

Problem 1

The critical t-value, as mentioned previously, would be $t^*=2.0096$ , making the 95% confidence interval equal to $CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})=390.25\pm2.0096(\frac{30.5}{\sqrt{50}})\approx\{381.5819,398.9181\}$

This interval suggests that we are 95% confident that the true mean levels of lead in soil are between 381.5819 and 398.9181 parts per million (ppm), which satisfies the EPA's regulated maximum of 400 ppm.

3 0

2 years ago

The cylinder C is being lifted using the cable and pulley system shown.

Sonja [21]

Answer:

sry but it's kinda hard

4 0

3 years ago

The idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse

sertanlavr [38]

Answer:

T = 5416.67 N

T = -2083.5 N

T = 0

Explanation:

Forward thrust has positive values and reverse thrust has negative values.

part a

Flight speed u = ( 150 km / h ) / 3.6 = 41.67 km / s

The thrust force represents the horizontal or x-component of momentum equation:

$T = flow(m_{exhaust})*(u_{exhaust} - u_{flight} )\\T = (50 kg/s ) * (150 - 41.67)\\\\T = 5416.67 N$

Answer: The thrust force T = 5416.67 N

part b

Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component, thus thrust equation is:

$T = flow(m_{exhaust})*(u_{exhaust} - u_{flight} )\\T = (50 kg/s ) * (0 - 41.67)\\\\T = -2083.5 N$

Answer: The thrust force T = -2083.5 N reverse direction

part c

Now the exhaust velocity and flight velocity is zero, then it has a zero horizontal component, thus thrust is also zero as there is no difference in two velocities in x direction.

Answer: T = 0 N

5 0

3 years ago

The following laboratory test results for Atterberg limits and sieve-analysis were obtained for an inorganic soil. [6 points] Si

alexira [117]

Answer: hello the complete question is attached below

answer:

A) Group symbol = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

Explanation:

A) Classifying the soil according to USCS system

( using 2nd image attached below )

description of sand :

The soil is a coarse sand since ≤ 50% particles are retained on No 200 sieve, also

The soil is a sand given that more than 50% particles passed from No 4 sieve

The soil can be a clean sand given that fines ≤ 12%

The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time

Group symbol as per the 2nd image attached below = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

5 0

3 years ago

What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 3 × 10-4

Vladimir [108]

Answer:

maximum stress is 2872.28 MPa

Explanation:

given data

radius of curvature = 3 × $10^{-4}$ mm

crack length = 5.5 × $10^{-2}$ mm

tensile stress = 150 MPa

to find out

maximum stress

solution

we know that maximum stress formula that is express as

$\sigma m = 2 ( \sigma o ) \sqrt{\frac{a}{\delta t}}$ ......................1

here σo is applied stress and a is half of internal crack and t is radius of curvature of tip of internal crack

so put here all value in equation 1 we get

$\sigma m = 2 ( \sigma o) \sqrt{\frac{a}{\delta t}}$

$\sigma m = 2(150) \sqrt{ \frac{\frac{5.5*10^{-2}}{2}}{3*10^{-4}}}$

σm = 2872.28 MPa

so maximum stress is 2872.28 MPa

8 0

3 years ago