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julia-pushkina [17]

2 years ago

13

8. What is the density of an object with a mass of 290.5 g and volume of 83 cm 3?

1 answer:

Leni [432]2 years ago

6 0

Answer:

The density would be 218.5

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Discuss the impact of the changing urban center. Include the impacts on political, economic, and social roles and opportunities.

KengaRu [80]

Answer:

The 21st century world have been earmarked with great influx of people to the urban centre,the notion of gender equality and female education have also made most traditional roles in the family changing.Before now,wives we're known for their full independence on their husband who is considered the bread winner.

Inspite of the growth of of the urban centre the availability of resources have dwindled,resulting in the surge of unemployment in many urban centre,the political entity of the society which is the government have serious challenging in managing the various threat posed by overpopulation, unemployment results in the decrease of standard of living of person and family,to cater for this family have to change their roles,wives now work to support the husband.

Explanation:

6 0

3 years ago

How does sea navigation work?

ahrayia [7]

Answer:

a clock

Explanation:

you use a clock in water

3 0

2 years ago

Which task best fits the role of a planning engineer?

Nonamiya [84]

Answer:

D

Explanation:

ensuring project end on time through carefully planning and organizing

8 0

3 years ago

At steady state, a reversible refrigeration cycle discharges energy at the rate QH to a hot reservoir at temperature TH, while r

ludmilkaskok [199]

Answer:

a) $COP_{R} = 25.014$ , b) $T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)$

Explanation:

a) The coefficient of performance of a reversible refrigeration cycle is:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

Temperatures must be written on absolute scales (Kelvin for SI units, Rankine for Imperial units)

$COP_{R} = \frac{275.15\,K}{286.15\,K-275.15\,K}$

$COP_{R} = 25.014$

b) The respective coefficient of performance is determined:

$COP_{R} = \frac{Q_{L}}{Q_{H}-Q_{L}}$

$COP_{R} = \frac{8.75\,kW}{10.5\,kW-8.75\,kW}$

$COP_{R} = 5$

But:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

The temperature at hot reservoir is found with some algebraic help:

$COP_{R} \cdot (T_{H}-T_{L})=T_{L}$

$T_{H}-T_{L} = \frac{T_{L}}{COP_{R}}$

$T_{H} = T_{L}\cdot \left(1+\frac{1}{COP_{R}} \right)$

$T_{H} = 273.15\,K \cdot \left(1+\frac{1}{5} \right)$

$T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)$

8 0

3 years ago

Read 2 more answers

At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power ge

tangare [24]

Answer:

e= 50 J/kg

Explanation:

Given that

Speed ,v= 10 m/s

Diameter of the turbine = 90 m

Density of the air ,ρ = 1.25 kg/m³

We know that mechanical energy given as

$E=\dfrac{1}{2}mv^2\ J$

That is why mechanical energy per unit mass will be

$e=\dfrac{1}{2}v^2\ J/kg$

Now by putting the values in the above equation we get

$e=\dfrac{1}{2}\times 10^2\ J/kg$

e= 50 J/kg

That why the mechanical energy unit mass will be 50 J/kg.

5 0

3 years ago