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Sav [38]
3 years ago
5

Consider the string length equal to 7. This string is distorted by a function f (x) = 2 sin(2x) - 10sin(10x). What is the wave f

ormed in this string? a. In=12cos (nit ) sin(max) b. 2cos(2t)sin (2x) - 10cos(10t ) sin(10x) c. n 2 sin 2x e' – 10sin 10x e
Engineering
1 answer:
fenix001 [56]3 years ago
8 0

Answer:

hello your question has a missing part below is the missing part

Consider the string length equal to \pi

answer : 2cos(2t) sin(2x) - 10cos(10t)sin(10x)

Explanation:

Given string length = \pi

distorted function f(x) = 2sin(2x) - 10sin(10x)

Determine the wave formed in the string

attached below is a detailed solution of the problem

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3 years ago
List six possible valve defects that should be included in the inspection of a used valve?
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Explanation:

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3 years ago
At an axial load of 22 kN, a 15-mm-thick × 40-mm-wide polyimide polymer bar elongates 4.1 mm while the bar width contracts 0.15
Alenkasestr [34]

Answer:

The Poisson's Ratio of the bar is 0.247

Explanation:

The Poisson's ratio is got by using the formula

Lateral strain / longitudinal strain

Lateral strain = elongation / original width (since we are given the change in width as a result of compession)

Lateral strain = 0.15mm / 40 mm =0.00375

Please note that strain is a dimensionless quantity, hence it has no unit.

The Longitudinal strain is the ratio of the elongation to the original length in the longitudinal direction.

Longitudinal strain = 4.1 mm / 270 mm = 0.015185

Hence, the Poisson's ratio of the bar is 0.00375/0.015185 = 0.247

The Poisson's Ratio of the bar is 0.247

Please note also that this quantity also does not have a dimension

3 0
3 years ago
A 0.19-m3 rigid tank equipped with a pressure regulator contains steam at 2 MPa and 300°C. The steam in the tank is now heated.
AVprozaik [17]

Answer:

576.21kJ

Explanation:

#We know that:

The balance mass m_{in}+m_{out}=\bigtriangleup m_{system}

so, m_e=m_1-m_2

Energy \ Balance\\E_{in}-E_{out}=\bigtriangleup E_{system}\\\\\therefore Q_i_n+m_eh_e=m_2u_2-m_1u_1

#Also, given the properties of water as;

(P_1=2Mpa,T_1=300\textdegree C)->v_1=0.12551m^3/kg,u_1=2773.2kJ/kg->h_1=3024.2kJ/kg\\\\(P_2=2Mpa,T_1=500\textdegree C)->v_2=0.17568m^3/kg,u_1=3116.9kJ/kg->h_1=3468.3kJ/kg

#We assume constant properties for the steam at average temperatures:h_e=\approx(h_1+h_2)/2

#Replace known values in the equation above;h_e=(3024.2+3468.3)/2=3246.25kJ/kg\\\\m_1=V_1/v_1=0.19m^3/(0.12551m^3/kg)=1.5138kg\\\\m_2=V_2/v_2=0.19m^3/(0.17568m^3/kg)=1.0815kg

#Using the mass and energy balance relations;

m_e=m_1-m_2\\\\m_e=1.5138-1.0815\\\\m_e=0.4323kg

#We have Q_i_n+m_eh_e=m_2u_2-m_1u_1: we replace the known values in the equation as;

Q_i_n+m_eh_e=m_2u_2-m_1u_1\\\\Q_i_n=0.4323kg\times3246.2kJ/kg+1.0815kg\times3116.9-1.5138kg\times2773.2kJ/kg\\\\Q_i_n=573.21kJ

#Hence,the amount of heat transferred when the steam temperature reaches 500°C is 576.21kJ

5 0
3 years ago
(25%) A well-insulated compressor operating at steady state takes in air at 70 oF and 15 psi, with a volumetric flow rate of 500
lubasha [3.4K]

Answer:

You can look it up

Explanation: if you don't know what it is look it up on .

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