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Ber [7]
1 year ago
13

the flow energy of 124 L/min of a fluid passing a boundary to a system is 108.5 kJ/min. Determine the pressure at this point

Physics
1 answer:
Andreyy891 year ago
5 0

Answer:

The pressure at this point is 0.875 mPa

Explanation:

Given that,

Flow energy = 124 L/min

Boundary to system P= 108.5 kJ/min

P=1.81\ kW

We need to calculate the pressure at this point

Using formula of pressure

P=F\times v

P=A_{1}P_{1}\times v_{1}

Here, A_{1}v_{1}=Q_{1}

Where, v = velocity

Put the value into the formula

1.81 =P_{1}\times0.124\times\dfrac{1}{60}

P_{1}=\dfrac{1.81\times60}{0.124}

P_{1}=875.80\ kPa

P_{1}=0.875\ mPa

Hence, The pressure at this point is 0.875 mPa

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The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve
Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

\varepsilon _t=\dfrac{D-d}{d}

\varepsilon _t=\dfrac{32-30}{30}

\varepsilon _t=0.0667

We know that

n=-\dfrac{\epsilon _t}{\varepsilon _{long}}

\varepsilon _{long}=-\dfrac{\epsilon _t}{n}

\varepsilon _{long}=-\dfrac{0.0667}{0.45}

\varepsilon _{long}=-0.148

So the axial pressure

P=E\times \varepsilon _{long}

P=5\times 0.148

P=740 KPa

The movement in the sleeve

\Delta =\varepsilon _{long}\times L

\Delta =0.148\times 50

Δ=7.4 mm

6 0
1 year ago
A (1.25+A) kg bowling ball is hung on a (2.50+B) m long rope. It is then pulled back until the rope makes an angle of (12.0+C)o
daser333 [38]

Answer:

F = 0.535 N

Explanation:

Let's use the concepts of energy, at the highest and lowest point of the trajectory

Higher

   Em₀ = U = mg y

Lower

    Em_{f} = K = ½ m v²

    Emo =Em_{f}

    mg y = ½ m v2

    v = √ 2gy

   y = L - L cos θ

  v = √ (2g L (1-cos θ))

Now let's use Newton's second law n at the lowest point where the acceleration is centripetal

     F = ma

     a = v² / r

In turning radius is the cable length r = L

    F = m 2g (1-cos θ)

Let's calculate

    F = 2  1.25  9.8 (1 - cos 12)

    F = 0.535 N

   

7 0
2 years ago
(a) Two point charges totaling 8.00 μC exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the
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A sleeping 68 kg man has a metabolic power of 79 w .
Lesechka [4]
 <span>65W * 8h * 3600s/h = 1.9e6 J = 447 Cal </span>
3 0
2 years ago
A lump of steel of mass 10kg at 627 degree Celsius is dropped in 100kg oil at 30 degree Celsius . the specific heat of steel And
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Answer:

700J

Explanation:

8 0
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