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Ilia_Sergeevich [38]

3 years ago

7

Problem a) – c): Use the method of joints, the method of sections, or both to solve the following trusses. Draw F.B.Ds for all y

our calculations. 1) Identify all the zero force members in the trusses; 2) Calculate the reactions at the supports; 3) Determine the internal forces for marked members in each truss; 4) Indicate your final results on the truss system.

1 answer:

Ghella [55]3 years ago

7 0

Answer:

This is confusing sorry

Explanation:

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Air expands through a turbine from 10 bar, 900 K to 1 bar, 500 K. The inlet velocity is small compared to the exit velocity of 1

Elis [28]

Answer:

- the mass flow rate of air is 7.53 kg/s

- the exit area is 0.108 m²

Explanation:

Given the data in the question;

lets take a look at the steady state energy equation;

m" = W" $_{cv$ / [ (h₁ - h₂ ) - $\frac{V_2^2}{2}$ ]

Now at;

T₁ = 900K, h₁ = 932.93 k³/kg

T₂ = 500 K, h₂ = 503.02 k³/kg

so we substitute, in our given values

m" = [ 3200 kW × $\frac{1\frac{k^3}{s} }{1kW}$ ] / [ (932.93 - 503.02 )k³/kg - $\frac{100^2\frac{m^2}{s^2} }{2}$ | $\frac{ln}{kg\frac{m}{s^2} }$ || $\frac{1kJ}{10^3N-m}$ | ]

m" = 7.53 kg/s

Therefore, the mass flow rate of air is 7.53 kg/s

now, Exit area A₂ = v₂m" / V₂

we know that; pv = RT

so

A₂ = RT₂m" / P₂V₂

so we substitute

A₂ = {[ $(\frac{8.314}{28.97}\frac{k^3}{kg.K})$ × $500 K$ × $(7.54 kg/s)$ ] / [(1 bar)(100 m/s )]} | $\frac{1 bar}{10N/m^2}$ || $10^3N.m/1k^3$

A₂ = 0.108 m²

Therefore, the exit area is 0.108 m²

8 0

3 years ago

il63 [147K]

Answer:

True

Explanation:

Dual home host - it is referred to as the firewall that is incorporated with two or more networks. out of these two networks, one is assigned to the internal network and the other is for the network. The main purpose of the dual-homed host is to ensure that no Internet protocol traffic is induced between both the network.

The most simple example of a dual-homed host is a computing motherboard that is provided with two network interfaces.

7 0

4 years ago

You are given a noninverting 741 op-amp with a dc-gain of 23.6 dB. The input signal to this amplifier is;Vin(t) = (0.18)∙cos(2π(

Vsevolod [243]

Answer:

Output voltage equation is $V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)$

Explanation:

Given:

dc gain $A = 23.6$ dB

Input signal $V_{in} (t) = 0.18 \cos (2\pi (57000)t +18.3)$

Now convert gain,

$A = 10^{\frac{23.6}{20} } = 15.13$

DC gain at frequency $f = 0$ is given by,

$A = \frac{V_{out} }{V_{in} }$

$V_{out} =AV_{in}$

$V_{out} = 15.13 \times 0.18 \cos (2\pi (57000)t +18.3)$

At zero frequency above equation is written as,

$V_{out} = 2.72 \times \cos 18.3$

$V_{out} = 2.72$

Now we write output voltage as input voltage,

$V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)$

Therefore, output voltage equation is $V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)$

7 0

3 years ago

Which option identifies what engineers will do to help in the following scenario?

Tasya [4]

Answer:

a

Explanation:

4 0

3 years ago

Read 2 more answers

Technician a s ays both an ohmmeter and a self-powered test light may be used to test for continuity. technician b says both may

amm1812

Both A and B technicians are correct because both might be used to test fuses, according to technician B.

<h3>What is continuity?</h3>

The behavior of a function at a certain point or section is described by continuity. The limit can be used to determine continuity.

From the question:

We can conclude:

The technician claims that you may check for continuity using both an ohmmeter and a self-powered test light. Both might be used to test fuses, according to technician B.

Thus, both A and B technicians are correct because both might be used to test fuses, according to technician B.

Technician A says both an ohmmeter and a self-powered test light may be used to test for continuity. Technician B says both may be used to test fuses. Who is correct?

Learn more about the continuity here:

brainly.com/question/15025692

#SPJ1

5 0

2 years ago