Help me with physics

For fully developed laminar pipe flow in a circular pipe, the velocity profile is u(r) = 2(1-r2 /R2 ) in m/s, where R is the inn

sdas [7]

Answer:

a) $v_{max} = 2\ \textup{m/s}$

b) $v_{avg} = 1\ \textup{m/s}$

c) Q = 1.256 × 10⁻³ m³/s

Explanation:

Given:

The velocity profile as:

$u(r) = 2(1-\frac{r^2}{R^2} )$

Now, the maximum velocity of the flow is obtained at the center of the pipe

i.e r = 0

thus,

$v_{max}=u(0) = 2(1-\frac{0^2}{R^2} )$

or

$v_{max} = 2\ \textup{m/s}$

Now,

$v_{avg} = \frac{v_{max}}{2}\ \textup{m/s}$

or

$v_{avg} = \frac{2}}{2}\ \textup{m/s}$

or

$v_{avg} = 1\ \textup{m/s}$

Now, the flow rate is given as:

Q = Area of cross-section of pipe × $v_{avg}$

or

Q = $\frac{\pi D^2}{4}\times v_{avg}$

or

Q = $\frac{\pi 0.04^2}{4}\times 1$

or

Q = 1.256 × 10⁻³ m³/s

7 0

3 years ago

10 The magnitude J of the current density in a certain lab wire with a circular cross section of radius R = 2.00 mm is given by

rewona [7]

Answer:

I = 0.002593 A = 2.593 mA

Explanation:

Current density = J = (3.00 × 10⁸)r² = Br²

B = (3.00 × 10⁸) (for ease of calculations)

The current through outer section is given by

I = ∫ J dA

The elemental Area for the wire,

dA = 2πr dr

I = ∫ Br² (2πr dr)

I = ∫ 2Bπ r³ dr

I = 2Bπ ∫ r³ dr

I = 2Bπ [r⁴/4] (evaluating this integral from r = 0.900R to r = R]

I = (Bπ/2) [R⁴ - (0.9R)⁴]

I = (Bπ/2) [R⁴ - 0.6561R⁴]

I = (Bπ/2) (0.3439R⁴)

I = (Bπ) (0.17195R⁴)

Recall B = (3.00 × 10⁸)

R = 2.00 mm = 0.002 m

I = (3.00 × 10⁸ × π) [0.17195 × (0.002⁴)]

I = 0.0025929449 A = 0.002593 A = 2.593 mA

Hope this Helps!!!

4 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

den301095 [7]

A: 0.7 -0.8 and 0hrs
B: 0.2 - 0.4 and 1.0 -1.1 hrs
C: 0-0.2 hrs and 0.8 - 1.0 hrs
D: 0.4 - 0.7 hrs

4 0

3 years ago

Read 2 more answers

The electric potential a distance r from a small charge is proportional to what power of the distance from the charge?

Gekata [30.6K]

Answer:

The power of the distance is -1.

Explanation:

The equation for the electric potential of a point charge is given by $V= k\frac{q}{r}$

where V is the electric potential, k is Coulomb's constant (it has a value of $9.0*10^{9}$ with units $\frac{Nm^{2} }{c^{2} }$ ), q is the electric charge of the small charge and r is the distance from the charge.

Now, the power of a number is how many times we multiply that number by itself; we see r appears only once in the equation. So we know the power is 1. But we can see in the equation that k and q are divided by r, which means r is the denominator. This means the power of r is negative (-).

Therefore, the power of r is -1.

5 0

3 years ago

The energy carried by sound waves is called -

goldenfox [79]

Answer:

kinetic mechanical energy

8 0

3 years ago

Read 2 more answers

Help me with physics​

Help me with physics