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3 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]3 years ago

8 0

Answer: SPL of Biggie = 85 dB

Explanation: Please find the attached files for the solution

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Global Courier Services will ship your package based on how much it weighs and how far you are sending the package. Packages abo

denis23 [38]

Answer:

The code will be:

#include <stdio.h>

#include <stdlib.h>

main () {

double weight, shippingCharge, rate, segments;

int distance;

printf("Enter the weight: \n");

scanf("%lf", &weight);

printf("Enter the distance: \n");

scanf("%i", &distance);

if (weight <= 10) {

printf("Rate is $3.00 \n");

rate = 3;

} else {

printf("Rate is $5.00 \n");

rate = 5;

}

if (distance % 500 == 0) {

segments = distance / 500;

} else {

segments = distance / 500 + 1;

}

shippingCharge = rate * segments;

if (distance >1000) {

shippingCharge = shippingCharge + 10;

}

printf("Your shipping charge is $%lf\n", shippingCharge);

system ("pause");

}

8 0

3 years ago

A composite wall is composed of 20 cm of concrete block with k = 0.5 W/m-K and 5 cm of foam insulation with k = 0.03 W/m-K. The

wariber [46]

Answer:

4.8°C

Explanation:

The rate of heat transfer through the wall is given by:

$q=\frac{Ak}{L}dT$

$\frac{q}{A}=\frac{k}{L}dT$

Assumptions:

1) the system is at equilibrium

2) the heat transfer from foam side to interface and interface to block side is equal. There is no heat retention at any point

3) the external surface of the wall (concrete block side) is large enough that all heat is dissipated and there is no increase in temperature of the air on that side

${k_{fi}= 0.03 W/m.K$

${L_{fi}= 5 cm = 0.05 m$

${T_{fi}= 25 \°C$

${k_{cb} = 0.5 W/m.K$

${L_{cb}= 20 cm = 0.20 m$

${T_{cb}= 0 \°C$

${T_{m}= ? \°C =$ temperature at the interface

Solving for ${T_{m}$ will give the temperature at the interface:

$\frac{q}{A}=\frac{k_{fi} }{L_{fi} }(T_{fi} -T_{m})=\frac{k_{cb} }{L_{cb} }(T_{m} -T_{cb})$

$\frac{0.03}{0.05 }(25 -T_{m})=\frac{0.5}{0.2}(T_{m} -0})$

$15 -0.6T_{m}=2.5T_{m}$

$3.1T_{m}=15$

$T_{m}=4.8$

3 0

3 years ago

A fluid at 300 K flows through a long, thin-walled pipe of 0.2-m diameter. The pipe is enclosed in a concrete casing that is of

andrew-mc [135]

Answer:

The correct answer is "1341.288 W/m".

Explanation:

Given that:

T₁ = 300 K

T₂ = 500 K

Diameter,

d = 0.2 m

Length,

l = 1 m

As we know,

The shape factor will be:

⇒ $SF=\frac{2 \pi l}{ln[\frac{1.08 b }{d} ]}$

By putting the value, we get

⇒ $=\frac{2 \pi l}{ln[\frac{1.08\times 1}{0.2} ]}$

⇒ $=3.7258 \ l$

hence,

The heat loss will be:

⇒ $Q=SF\times K(T_2-T_1)$

$=3.7258\times 1\times 1.8\times (500-300)$

$=3.7258\times 1.8\times (200)$

$=1341.288 \ W/m$

3 0

3 years ago

Just need someone to talk to pls dont just use me for points

Allushta [10]

Answer:

well what do you wanna talk about friend?

Explanation:

7 0

3 years ago

The density of a certain type of steel is 8.1 g/cm3. What is the mass of a 100 cm3 chunk of this steel

irina1246 [14]

Answer:

810 g

Explanation:

Mass is the product of density and volume:

m = ρV

m = (8.1 g/cm³)(100 cm³) = 810 g

The mass of the chunk is 810 grams.

4 0

2 years ago