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My name is Ann [436]
3 years ago
10

The velocity profile in fully developed laminar flow in a circular pipe of inner radius R 5 2 cm, in m/s, is given by u(r) 5 4(1

2 r2/R2). Determine the average and maximum velocities in the pipe and the volume flow rate.
Physics
1 answer:
xxMikexx [17]3 years ago
8 0

The question is not clear and the complete clear question is;

The velocity profile in fully developed laminar flow In a circular pipe of inner radius R = 2 cm, in m/s, is given By u(r) = 4(1 - r²/R²). Determine the average and maximum Velocities in the pipe and the volume flow rate.

Answer:

A) V_max = 4 m/s

B) V_avg = 2 m/s

C) Flow rate = 0.00251 m³/s

Explanation:

A) We are given that;

u(r) = 4(1 - (r²/R²))

To obtain the maximum velocity, let's apply the maximum condition for a single-variable continual real valued problem to obtain;

(d/dr)(u(r)) = 0

Thus,

(d/dr)•4(1 - (r²/R²)) = 0

4(d/dr)(1 - (r²/R²)) = 0

If we differentiate, we have;

4(0 - (2r/R²)) = 0

-8r/R² = 0

Thus, r = 0 and with that, the maximum velocity is at the centre of the pipe.

Thus, for maximum velocity, let's put 0 for r in the U(r) function.

Thus,

V_max = 4(1 - 0²/R²) = 4 - 0 = 4 m/s

B) Average velocity is given by;

V_avg = V_max/2

V_avg = 4/2 = 2 m/s

C) the flow can be calculated from;

Flow rate ΔV = A•V_avg

A is area = πr²

From question, r = 2cm = 0.02m

A = π x 0.02²

Hence,

ΔV = π x 0.02² x 2 = 0.00251 m³/s

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Answer:

Width of the slit will be equal to 1.47 mm

Explanation:

We have given wavelength of the light \lambda =550nm=550\times 10^{-9}m

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So x=3\times 10^{-3}m

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So width of the slit will be equal to 1.47 mm

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A motorcycle of mass 250 kg drives around a circle with a centripetal acceleration of 3.4 m/s2. What is the centripetal force ac
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A lemon with mass 0.3 kg falls out of a tree from a height of 1.8 m. How much mechanical energy does the lemon have just before
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Mechanical energy is the energy that is possessed by an object due to its motion or due to its position. It can either be kinetics or potential. In this problem you know it starting position so you can calculate it's potential energy (PE):

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A projectile is fired into the air from the top of a 200-m cliff above a valley as shown below. Its initial velocity is 60 m/s a
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a) y(max)  = 337.76 m

b) t₁ = 5.30 s  the time for y maximum

c)t₂ =  13.60 s  time for y = 0 time when the fly finish

d) vₓ = 30 m/s        vy = - 81.32 m/s

e)x = 408 m

Equations for projectile motion:

v₀ₓ = v₀ * cosα          v₀ₓ = 60*(1/2)     v₀ₓ = 30 m/s   ( constant )

v₀y = v₀ * sinα           v₀y = 60*(√3/2)     v₀y = 30*√3  m/s

a) Maximum height:

The following equation describes the motion in y coordinates

y  =  y₀ + v₀y*t - (1/2)*g*t²      (1)

To find h(max), we need to calculate t₁ ( time for h maximum)

we take derivative on both sides of the equation

dy/dt  = v₀y  - g*t

dy/dt  = 0           v₀y  - g*t₁  = 0    t₁ = v₀y/g

v₀y = 60*sin60°  = 60*√3/2  = 30*√3

g = 9.8 m/s²

t₁ = 5.30 s  the time for y maximum

And y maximum is obtained from the substitution of t₁  in equation (1)

y (max) = 200 + 30*√3 * (5.30)  - (1/2)*9.8*(5.3)²

y (max) = 200 + 275.40 - 137.64

y(max)  = 337.76 m

Total time of flying (t₂)  is when coordinate y = 0

y = 0 = y₀  + v₀y*t₂ - (1/2)* g*t₂²

0 = 200 + 30*√3*t₂  - 4.9*t₂²            4.9 t₂² - 51.96*t₂ - 200 = 0

The above equation is a second-degree equation, solving for  t₂

t =  [51.96 ±√ (51.96)² + 4*4.9*200]/9.8

t =  [51.96 ±√2700 + 3920]/9.8

t =  [51.96 ± 81.36]/9.8

t = 51.96 - 81.36)/9.8         we dismiss this solution ( negative time)

t₂ =  13.60 s  time for y = 0 time when the fly finish

The components of the velocity just before striking the ground are:

vₓ = v₀ *cos60°       vₓ = 30 m/s  as we said before v₀ₓ is constant

vy = v₀y - g *t        vy = 30*√3  - 9.8 * (13.60)

vy = 51.96 - 133.28         vy = - 81.32 m/s

The sign minus means that vy  change direction

Finally the horizontal distance is:

x = vₓ * t

x = 30 * 13.60  m

x = 408 m

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2 years ago
The magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same
horrorfan [7]

Answer:

2653 turns

Explanation:

We are given that

Diameter,d=2 cm

Length of magnet,l=8 cm=8\times 10^{-2} m

1m=100 cm

Magnetic field,B=0.1 T

Current,I=2.4 A

We are given that

Magnetic field of solenoid and magnetic are same and size of both solenoid and magnetic are also same.

Length of solenoid=8\times 10^{-2} m

Magnetic field of solenoid

B=\frac{\mu_0NI}{l}

Using the formula

0.1=\frac{4\pi\times 10^{-7}\times 2.4\times N}{8\times 10^{-2}}

Where \mu_0=4\pi\times 10^{-7}

N=\frac{0.1\times 8\times 10^{-2}}{4\pi\times 10^{-7}\times 2.4}=2653 turns

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3 years ago
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