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NeTakaya
3 years ago
9

Air flows into a jet engine at 70 lbm/s, and fuel also enters the engine at a steady rate. The exhaust gases, having a density o

f 0.1 lbm/ft3 , exit through a circular cross section with a radius of 1 ft at 1450 ft/s relative to the engine. Find the mass of fuel which is supplied to the engine each minute.
Physics
1 answer:
Andrews [41]3 years ago
8 0

Answer:

1387908 lbm/h

Explanation:

Air flowing into jet engine = 70 lbm/s

ρ = Exhaust gas density = 0.1 lbm/ft³

r = Radius of exit with a circular cross section = 1 ft

v = Exhaust gas velocity = 1450 ft/s

Exhaust gas mass (flow rate)= Air flowing into jet engine + Fuel

Q = (70+x) lbm/s

Area of exit with a circular cross section = π×r² = π×1²= π m²

Now from energy balance

Q = ρ×A×v

⇒70+x = 0.1×π×1450

⇒70+x = 455.53

⇒ x = 455.53-70

⇒ x = 385.53 lbm/s

∴ Mass of fuel which is supplied to the engine each minute is 1387908 lbm/h

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the density of a steel is 7.8.find the mass of steel cube of side 10 cm. find volume of steel if the mass is 8kg​
lukranit [14]

Answer:

Mass of the steel cube =  7800 kg

Volume of the steel  = 1.025 cubic centimetre

Explanation:

Given:

The density of the steel   =  7.8

Side of the cube = 12 cm

<u>(1)The mass of steel cube :</u>

We know that,

Density  = \frac{Mass}{Volume}

We are given with density and sides of the cube

then volume of the cube

= (side)^3

= 10^3

= 1000 cubic centimetre

Now

7.8 = \frac{mass}{1000}

mass = 7.8 \times 1000

mass =  7800 kg

<u>(2)volume of steel:</u>

Given the mass  = 8 kg

Density  = \frac{Mass}{Volume}

Substituting the values

7.8 = \frac{8}{volume}

volume = \frac{8}{7.8}

volume = 1.025 cubic centimetre

3 0
3 years ago
Water initially at 200 kPa and 300°C is contained in a piston–cylinder device fitted with stops. The water is allowed to cool at
netineya [11]

Answer:

Δu=1300kJ/kg  

Explanation:

Energy at the initial state

p_{1}=200kPa\\t_{1}=300^{o}\\u_{1}=2808.8kJ/kg(tableA-5)

Is saturated vapor at initial pressure we have

p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)

Process 2-3 is a constant volume process

p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)

The overall in internal energy

Δu=u₁-u₃

We replace the values in equation

Δu=u₁-u₃

=2808.8kJ/kg-1508.6kJ/kg\\=1300kJ/kg

Δu=1300kJ/kg  

3 0
3 years ago
The summer camps had a field trip from the campus to Fragrance Hill. They traveled at an average speed of 65 km/h in the first 2
ludmilkaskok [199]

Answer:

Explanation:

They traveled this distance in 2 parts, essentially. Part 1 had an average speed for a certain number of hours, part 2 had an average speed for a certain number of hours, and those 2 parts taken together took them a distance of 364 km. In equation form, that looks like this:

km/hr part 1 + km/hr part 2 = 364 km

Now we need to find each part on the left side of that equation. Part 1 first:

We traveled 65 km/hr for 2 hours, so that took us

65\frac{km}{hr}*2hr and canceling out the hour label, we have that in part 1 we got

65(2) = 130 km. Good. Now onto the second part, where our unknown is.

We traveled 78 km/hr the second part for x hours, so that took us

78\frac{km}{hr}*xhr and canceling out the hour label, we have that in part 2 we got

78x km. Now we can fill in the main equation (the one in bold print)

130 km + 78x km = 364 km and subtracting 130 km from both sides:

78x km = 234 km and dividing by 78 km:

x = 3 hours. Part 2 took 3 hours. Part 1 took 2 hours, so the whole trip took 5 hours.

7 0
3 years ago
Zamir and Talia raced through a maze. Zamir walked 2 m north, 2 m east, 4 m south, 2 m east, 4 m north, 2 m east, 3 m south, 4 m
Alja [10]

Answer : Zamir's displacement and Talia's displacement is equal.

Explanation :

Displacement is explained to be the changing position of an object.

Zamir covers total distance 27 m and Talia covers total distance 19 m but  Zamir's initial and final position and Talia's initial and final position is same.

So, we can say that Zamir's displacement and Talia's displacement is equal.

6 0
3 years ago
Read 2 more answers
Assuming that Albertine's mass is 60.0 kg , for what value of μk, the coefficient of kinetic friction between the chair and the
makvit [3.9K]

Answer:

\mu_k=0.101

Explanation:

It is given that,

Mass of Albertine, m = 60 kg

It can be assumed, the spring constant of the spring, k = 95 N/m

Compression in the spring, x = 5 m

A glass sits 19.8 m from her outstretched foot, h = 19.8 m

When she just reach the glass without knocking it over, a force of friction will also act on it. Using the conservation of energy for the spring mass system such that,

\dfrac{1}{2}kx^2=\mu_k mgh

\mu_k=\dfrac{kx^2}{2mgh}

\mu_k=\dfrac{95\times (5)^2}{2\times 60\times 9.8\times 19.8}

\mu_k=0.101

So, the coefficient of kinetic friction between the chair and the waxed floor is 0.101. Hence, this is the required solution.

3 0
3 years ago
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