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svetoff [14.1K]
3 years ago
11

The 5.5 million vinyl long-playing (LP) records sold in the United States per year pales in comparison with the 1.26 billion dig

ital downloads. The grooves in vinyl records create sound waves that are similar to those of a live performance, however, which means they provide a more authentic sound, which in turn means nightclub DJs, discerning music listeners, and collectors will always prefer them. Firms that sell LPs position themselves for a niche segment of diehard consumers. They are in the ________ stage.
Physics
1 answer:
miv72 [106K]3 years ago
4 0

Answer:

decline

Explanation:

Based on the scenario being described within the question it can be said that these types of firms are in the decline stage of the product life cycle. This stage refers to when a product has already passed it's peak potential and sales begin to decline until production is ultimately halted and the product dies off. Which is exactly what is happening to the LP's since everyone has moved on to digital downloads.

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A driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di
Oksi-84 [34.3K]
<h2>82.353 km/hr</h2>

Explanation:

       The driver travels 135 km towards East in 1.5 hr. He stops for 45 min. He again travels 215 km towards East in 2.0 hr.

       The total displacement of the driver in the given time is ths sum of individual displacements, because all the displacements are in the same directon.

       Total displacement = 135\text{ }km\text{ }East\text{ }+\text{ }0\text{ }km\text{ }+\text{ }215\text{ }km\text{ }East\text{ }=\text{ }350\text{ }km\text{ }East

       Total time travelled = 1.5\text{ }hr\text{ }+\text{ }45\text{ }min\text{ }+\text{ }2.0\text{ }hr\text{ }=\text{ }90\text{ }min\text{ }+\text{ }45\text{ }min\text{ }+\text{ }120\text{ }min\text{ }=\text{ }255\text{ }min\text{ }=\text{ }4\text{ }hr\text{ }15\text{ }min\text{ }=\text{ }4.25\text{ }hr

      \text{Average velocity = }\dfrac{\text{Total displacement}}{\text{Total time taken}}=\dfrac{350\text{ }km}{4.25\text{ }hr}=82.353\text{ }\frac{km}{hr}

∴ Driver's average velocity = 82.353\text{ }\frac{km}{hr}

3 0
3 years ago
A 5.67-kg block of wood is attached to a spring with a spring constant of 150 N/m. The block is free to slide on a horizontal fr
nikklg [1K]

Answer:

v=0.94 m/s

Explanation:

Given that

M= 5.67 kg

k= 150 N/m

m=1 kg

μ = 0.45

The maximum acceleration of upper block can be μ g.

 a= μ g                          ( g = 10 m/s²)

The maximum acceleration of system will ω²X.

ω = natural frequency

X=maximum displacement

For top stop slipping

μ g  =ω²X

We know for spring mass system natural frequency given as

\omega=\sqrt{\dfrac{k}{M+m}}

By putting the values

\omega=\sqrt{\dfrac{150}{5.67+1}}

ω = 4.47 rad/s

μ g  =ω²X

By putting the values

0.45 x 10 = 4.47² X

X = 0.2 m

From energy conservation

\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2

kX^2=(m+M)v^2

150 x 0.2²=6.67 v²

v=0.94 m/s

This is the maximum speed of system.

7 0
3 years ago
Read 2 more answers
A 1-kg iron frying pan is placed on a stove. The pan increases from 20°C to 250°C. If the same amount of heat is added to a pan
Nesterboy [21]

Answer;

The temperature change for the second pan will be lower compared to the temperature change of the first pan

Explanation;

-The quantity of heat is given by multiplying mass by specific heat and by temperature change.

That is; Q = mcΔT

This means; the quantity of heat depends on the mass, specific heat capacity of a substance and also the change in temperature.

-Maintaining the same quantity of heat, with another pan of the same mass and greater specific heat capacity would mean that the change in temperature would be much less lower.

7 0
3 years ago
What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.
Studentka2010 [4]

<u>Answer:</u> The change in entropy of the given process is 1324.8 J/K

<u>Explanation:</u>

The processes involved in the given problem are:

1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})      .......(1)

where,

\Delta S = Entropy change

C_{p,m} = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g    (Conversion factor: 1 kg = 1000 g)

T_2 = final temperature

T_1 = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=m\times \frac{\Delta H_{f,v}}{T}      .......(2)

where,

\Delta S = Entropy change

m = mass of ice

\Delta H_{f,v} = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

  • <u>For process 1:</u>

We are given:

m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K

Putting values in equation 1, we get:

\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K

  • <u>For process 2:</u>

We are given:

m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K

  • <u>For process 3:</u>

We are given:

m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K

Putting values in equation 1, we get:

\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K

  • <u>For process 4:</u>

We are given:

m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K

  • <u>For process 5:</u>

We are given:

m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K

Putting values in equation 1, we get:

\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K

Total entropy change for the process = \Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5

Total entropy change for the process = [21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K

Hence, the change in entropy of the given process is 1324.8 J/K

4 0
3 years ago
Is cutting a string physcal or chemical change
oksano4ka [1.4K]
This is a physical change because cutting the string didn't change it chemically, but it did physically.
6 0
3 years ago
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