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2 years ago

9

A family quarantined at home in March/April 2020 has two dogs: a bull mastiff (Biggie), and a chihuahua (Smalls). Smalls has a b

ark with SPL of 80 dB. When the mail carrier comes to drop off the mail, both Biggie and Smalls bark and the SPL of their combined barks is 90 dB. What is the SPL of Biggie?

1 answer:

liubo4ka [24]2 years ago

8 0

Answer: SPL of Biggie = 85 dB

Explanation: Please find the attached files for the solution

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Una frase de: ama la vida quien___________________________________

TEA [102]

Answer:

A phrase from: who loves life

Explanation:

5 0

2 years ago

A counter-flow double pipe heat exchanger is heat heat water from 20 degrees Celsius to 80 degrees Celsius at the rate of 1.2 kg

lakkis [162]

Answer:

$L=107.6m$

Explanation:

Cold water in: $m_{c}=1.2kg/s, C_{c}=4.18kJ/kg\°C, T_{c,in}=20\°C, T_{c,out}=80\°C$

Hot water in: $m_{h}=2kg/s, C_{h}=4.18kJ/kg\°C, T_{h,in}=160\°C, T_{h,out}=?\°C$

$D=1.5cm=0.015m, U=649W/m^{2}K, LMTD=?\°C, A_{s}=?m^{2},L=?m$

Step 1: Determine the rate of heat transfer in the heat exchanger

$Q=m_{c}C_{c}(T_{c,out}-T_{c,in})$

$Q=1.2*4.18*(80-20)$

$Q=1.2*4.18*(80-20)$

$Q=300.96kW$

Step 2: Determine outlet temperature of hot water

$Q=m_{h}C_{h}(T_{h,in}-T_{h,out})$

$300.96=2*4.18*(160-T_{h,out})$

$T_{h,out}=124\°C$

Step 3: Determine the Logarithmic Mean Temperature Difference (LMTD)

$dT_{1}=T_{h,in}-T_{c,out}$

$dT_{1}=160-80$

$dT_{1}=80\°C$

$dT_{2}=T_{h,out}-T_{c,in}$

$dT_{2}=124-20$

$dT_{2}=104\°C$

$LMTD = \frac{dT_{2}-dT_{1}}{ln(\frac{dT_{2}}{dT_{1}})}$

$LMTD = \frac{104-80}{ln(\frac{104}{80})}$

$LMTD = \frac{24}{ln(1.3)}$

$LMTD = 91.48\°C$

Step 4: Determine required surface area of heat exchanger

$Q=UA_{s}LMTD$

$300.96*10^{3}=649*A_{s}*91.48$

$A_{s}=5.07m^{2}$

Step 5: Determine length of heat exchanger

$A_{s}=piDL$

$5.07=pi*0.015*L$

$L=107.57m$

7 0

2 years ago

(a) Differentiate between heat treatment of ferrous and non-ferrous alloys (b) With your understanding of material's thermal pro

liubo4ka [24]

Answer:

In ferrous metal iron present but on the other hand in the non ferrous material iron does not present.That is why there is a different heat treatment process for ferrous and nonferrous materials.

Ferrous materials contains iron is the main constitute.Like steel ,cast iron ,wrought iron .Steel and cast iron are the alloy element of iron ans carbon.Wrought iron is the purest from of iron.

Heat treatment process for ferrous materials :

1.Normalizing

2.Annealing

3.Quenching

4.Surface hardening

Heat treatment process for non ferrous materials :

Mostly annealing process is used for non ferrous materials.After annealing non ferrous will become soft.

When two metal plates are joined then they form a bimetallic structure.The bimetallic structure is used to find the relationship of thermal temperature and the mechanical displacement.

The use of bimetallic structure -In clock ,thermometers ,engines.

7 0

3 years ago

Student A says hazardous waste can take the form of solid, liquid, or gas. Student B says hazardous waste can only take the form

lina2011 [118]

Answer:

Student A

Explanation:

hope this helps have a great day

4 0

3 years ago

A diagonal aluminum alloy tension rod of diameter d and initial length l is used in a rectangular frame to prevent collapse. The

Ksenya-84 [330]

Answer:

$\Delta L = 0.1883 inch$

Explanation:

Given data:

d = 0.5 inch

$L_i = 8 ft$

$\sigma_{allow} = 20 kpsi$

we know that change in length is given as

$\Delta L = \frac{PL}{AE}$

$= \frac{P}{A}\times \frac{L}{E}$

$= \sigma_{allow} \times \frac{L}{E}$

modulus of elasticity E for aluminium alloy is $10.2 \times 10^6 psi = 10.2 \times 10^3 kpsi$

$\Delta L = \frac{20 \times 8}{10.2\times 10^3}$

$\Delta L = 0.01569 ft$

$\Delta L = 0.1883 inch$

6 0

3 years ago