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3 years ago

14

Water flows through a horizontal pipe. The diameter of the pipe at point b is larger than the diameter of the pipe at point a. W

here is the speed of the water the greatest?

1 answer:

Natalija [7]3 years ago

5 0

The speed of the water is the greatest at point B

You might be interested in

A solenoid is designed to produce a magnetic field of 3.50×10^−2 T at its center. It has a radius of 1.80 cm and a length of 46.

Anna11 [10]

Answer:

a. 2143 turns/m

b. 111.5 m

Explanation:

a. The minimum number of turns per unit length (N/L) can be found using the following equation:

$B = \frac{\mu_{0}NI}{L}$

$\frac{N}{L} = \frac{B}{\mu_{0}I} = \frac{3.50 \cdot 10^{-2} T}{4\pi \cdot 10^{-7} Tm/A*13.0 A} = 2143 turns/m$

Hence, the minimum number of turns per unit length is 2143 turns/m.

b. The total length of wire is the following:

$N = 2143 turns/m*L = 2143 turns/m*46.0 \cdot 10^{-2} m = 986 turns$

Since each turn has length 2πr of wire, the total length is:

$L_{T} = N*2\pi r = 986 turn*2*\pi*1.80 \cdot 10^{-2} m = 111.5 m$

Therefore, the total length of wire required is 111.5 m.

I hope it helps you!

4 0

3 years ago

A man is standing on a weighing machine on a ship which is bobbing up and down with simple harmonic motion of period T=15.0s.Ass

STALIN [3.7K]

Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force. We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.

If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as 588 newtons or as
132.3 pounds. That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.

If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is

   y(t) = y₀ + M sin(2π t/15) .

The vertical speed of the deck is   y'(t) = M (2π/15) cos(2π t/15)

and its vertical acceleration is y''(t) = - (2πM/15) (2π/15) sin(2π t/15)

      = - (4 π² M / 15²) sin(2π t/15)

      = - 0.1755 M sin(2π t/15) .

There's the important number ... the 0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.

The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of 0.1755 x amplitude).

At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of 65kg, when in reality it's only 60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.

Now I'm going to wave my hands in the air a bit:

Apparent weight = (apparent mass) x (real acceleration of gravity)

(Apparent mass) = (65/60) = 1.08333 x real mass.

Apparent 'gravity' = 1.08333 x real acceleration of gravity.

The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.

0.08333 G = 0.1755 M

The 'M' is what we need to find.

Divide each side by 0.1755 : M = (0.08333 / 0.1755) G

'G' = 9.0 m/s²
   M = (0.08333 / 0.1755) (9.8) = 4.65 meters .

That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .

8 0

3 years ago

What is the eulerian description of fluid motion how does it differ from the lagrangian description?

Alex_Xolod [135]

Kinematics : Study of motion

Fluid kinematics : study of how fluid flows and how to describe its motion.

There are two ways to describe fluid motion

one is Eulerian, where the variations are described at all fixed stations as a function of time.

the other is Lagrangian, in which one follows all fluid particles and describes the variations around each fluid particle along its trajectory.

<u>DIFFRENCE BETWEEN LAGRANGIAN AND EULERIAN:</u>

1.Both Lagrangian and Eulerian describes time variation.

2. Eulerian describes the rate of change in one point of space

Lagrangian descries rate of change of a property of material system.

To know more about the Lagrangian and Eulerian :\brainly.com/question/14944792

#SPJ4

3 0

2 years ago

Mixing of sand in water is an irreversible change true aur false

USPshnik [31]

Answer:

<h3>true</h3>

Explanation:

<h3>hope it helps you ❤️</h3><h3>happy to help</h3>

3 0

2 years ago

A parallel-plate capacitor has square plates that are 7.20 cm on each side and 3.40 mm apart. The space between the plates is co

Lilit [14]

Answer:

U = 218 nJ

Explanation:

We are given;

Spacing between the plates; d = 3.4 mm = 3.4 × 10^(-3) m

Voltage across the capacitor; V = 96 V

Dimension of the square plates is 7.2cm x 7.2cm.

So, Area = 7.2 × 7.2 = 51.84 cm² = 51.84 × 10^(-4) m²

Permittivity of free space; ε_o = 8.85 × 10^(-12) C²/N.m²

From relative permeability table;

Dielectric constant of Pyrex; k1 = 5.6

Dielectric constant of polystyrene; k2 = 2.56

Now, formula for capacitance of a capacitor with Dielectric is;

C = kC_o

Where, C_o = ε_o(A/d)

Since there are 2 capacitors, d will now be d/2 = (3.4 × 10^(-3))/2 m = 1.7 × 10^(-3)

Since we have 2 capacitor, thus ;

C1 = k1*ε_o*(A/d)

C1 = (5.6 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))

C1 = 1.51 × 10^(-10) F

Similarly;

C2 = (2.56 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))

C2 = 0.691 × 10^(-10) F

For capacitors in series, formula for total capacitance(Cs) is;

1/Cs = (1/C1) + (1/C2)

Simplifying this, we have;

Cs = (C1*C2)/(C1 + C2)

Plugging in the relevant values ;

Cs = (1.51 × 10^(-10)*0.691 × 10^(-10))/((1.51 × 10^(-10)) + (0.691 × 10^(-10)))

Cs = 0.474 × 10^(-10) F

The formula for energy stored in a capacitor with 2 Dielectrics is given as;

U = ½Cs*V²

So,

U = ½ × 0.474 × 10^(-10) × 96²

U = 2.18 × 10^(-7) J = 218 × 10^(-9) = 218 nJ

8 0

3 years ago