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wolverine [178]
3 years ago
13

Air enters a nozzle steadily at 2.21 kg/m3 and 40 m/s and leaves at 0.752 kg/m3 and 180 m/s. If the inlet area of the nozzle is

90 cm2, determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle.
(a) 0.796 kg/s,
(b) 58 cm2
Physics
1 answer:
Anestetic [448]3 years ago
7 0

Answer

Given,

Air enter density, ρ₁ = 2.21 kg/m³

speed of entry, v₁ = 40 m/s

Air exit density , ρ₂ = 0.752 kg/m³

speed of exit, v₂ = 180 m/s

Inlet area = 90 cm²

a) mass flow rate through nozzle.

   m = \rho_1 A_1 v_1

   m = 2.21\times 90\times 10^{-4}\times 40

            m = 0.796 kg/s

b) exit area

  Using continuity equation

  \rho_1 A_1 v_1=\rho_2A_2 v_2

  A_2 = \dfrac{\rho_1 A_1 v_1}{\rho_2 v_2}

  A_2 = \dfrac{2.21\times 90\times 40}{0.752\times 180}

  A_2 = 58\ cm^2

Exit area of nozzle is equal to 58\ cm^2

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ivanzaharov [21]

Answer:

a)

Explanation:

Newton's First law (aka law of inertia) says that any object, not subject to an external influence, must be keep in its present motion status (at rest or moving at constant speed along a straight line) forever.

So, if we see an object in motion slowing down, there must be an external infuence acting on it, that makes it slow down, like a friction force, for instance.

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A conductor carrying 14.7 amps of current is directed along the positive x-axis and perpendicular to a uniform magnetic field. A
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Explanation:

It is given that,

Current carried in the conductor, I = 14.7 A (+x axis)

The magnetic force per unit length on the conductor, \dfrac{F}{L}=0.125\ N/m (-y axis)

The magnetic force is given by :

F=ILBsin\theta

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F=ILB

B=\dfrac{F}{IL}

B=\dfrac{0.125}{14.7}  

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4 0
3 years ago
We often see cracks on rocks and stones why
Paul [167]

Answer:

Due to the process of weathering and erosion.

Explanation:

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  • The process of physical weathering that is called as mechanical weathering of rocks tends to break them apart and leads to the development of crack and small openings.
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The kinetic energy of a moving object is E=12mv2. A 61 kg runner is moving at 10kmh. However, her speedometer is only accurate t
jek_recluse [69]

Answer:

e=3367.2J

%e=1.43%

Explanation:

From the exercise we know two information. The real speed and the experimental measured by the speedometer

v_{r}=10km/h=2.77m/s

Since the speedometer is only accurate to within 0.1km/h the experimental speed is

v_{e}=10km/h-0.1km/h=9.9km/h=2.75m/s

Knowing that we can calculate Kinetic energy for the real and experimental speed

E_{r}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.77m/s)^2=234023J

E_{e}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.75m/s)^2=230656J

Now, the potential error in her calculated kinetic energy is:

e=E_{r}-E_{e}=(234023-230656)J=3367.2J

%e=\frac{E_{r}-E_{e}}{E_{r}}x100=\frac{(234023-230656)J}{234023J}x100=1.43%

4 0
4 years ago
9. What is the total kinetic energy (KEtran + KErot) of a solid cylinder with mass M = 2.50 kg and radius R = 0.5 m that rolls w
GarryVolchara [31]

Answer:

110.25 J

Explanation:

We are given that

Mass,M=2.5 kg

Radius,R=0.5 m

Distance,d=4.5 m

Initial speed,u=0

We have to find the total kinetic energy

According to law of conservation of energy

Total kinetic energy=Potential energy=mgh

Where g=9.8m/s^2

Using the formula

Total kinetic energy=2.5\times 9.8\times 4.5

Total kinetic energy=110.25 J

4 0
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