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1 year ago

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A viscous liquid is sheared between two parallel disks of radius �, one of which rotates with angular speed Ω, while the other i

s fixed. The velocity field is purely tangential, and the velocity varies linearly with z from: �; = 0 at � = 0 (the fixed disk) to the velocity of the rotating disk at its surface (� = ℎ). Derive an expression for the velocity field between the disks.

1 answer:

Alexus [3.1K]1 year ago

5 0

Answer:

Upper disk rotates at a constant angular velocity. The velocity at any height from stationery disk, say at x metres

$U_o=v(\frac {x}{h})$ where v is tangential velocity at radius r from the centre of disk

$U_o=r\omega (\frac {x}{h})$

The radial component of velocity is given as

$U_r=0$

The z component of velocity is also given as

W=0

Total velocity, $v= r\omega (\frac {x}{h})\hat e_{o}$

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An antibaryon composed of two antiup quarks

Sveta_85 [38]

Answer:

(2) −1 e

Explanation:

A quark is the lightest elementary particles which form hadron such as proton and neutron. A quark has fractional charge.

Up, charm and top quarks have $+\frac{2}{3} e$ charge where as down, strange and bottom quarks have $-\frac{1}{3}e$ charge.

The antiparticle of up quark is antiup quark and has charge $-\frac{2}{3}e$ charge.

The antiparticle of down quark is antidown quark and has charge $+\frac{1}{3}e$ charge.

An antibaryon is composed of two anti-up quark and one anti-down quark.

Net charge of the anti-baryon is:

$2\times (-\frac{2}{3} e)+1\times (+\frac{1}{3})e=-1e$

Thus, antibaryon has -1e charge.

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2 years ago

A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.

Sedaia [141]

Answer:

335°C

Explanation:

Heat gained or lost is:

q = m C ΔT

where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Heat gained by the water = heat lost by the copper

mw Cw ΔTw = mc Cc ΔTc

The water and copper reach the same final temperature, so:

mw Cw (T - Tw) = mc Cc (Tc - T)

Given:

mw = 390 g

Cw = 4.186 J/g/°C

Tw = 22.6°C

mc = 248 g

Cc = 0.386 J/g/°C

T = 39.9°C

Find: Tc

(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)

Tc = 335

7 0

1 year ago

While practicing S-turns, a consistently smaller half-circle is made on one side of the road than on the other, and this turn is

erica [24]

Answer:

The answer is "4-5-6 because the bank is growing too quickly in the beginning of its turn".

Explanation:

The S-turns is also known as the reference technique, in which the ground track of the aircraft on both sides of a defined ground-based straight-line distance represents 2 different but equivalent circles.

Throughout the S-turns, on either side of the road, a progressively smaller half-circle is formed, and this turn does not stop until the road or reference line is crossed during the early part of the turn, the twists increase too quickly.

5 0

1 year ago

A) The current theory of the structure of the Earth, called plate tectonics, tells us that the continents are in constant motion

suter [353]

A) The mass of the continent is $2.5\cdot 10^{21} kg$

B) The kinetic energy is 2016 J

C) The speed of the jogger should be 7.1 m/s

Explanation:

A)

The mass of the continent can be calculated as

$m = \rho V$

where

$\rho = 2800 kg/m^3$ is its density

V is its volume

We have to calculate its volume. We know that the continent is represented as a slab of side 5900 km (so its surface is 5900 x 5900, assuming it is a square) and depth of 26 km, so its volume is:

$V=(5900 km)^2 (26 km)=9.05\cdot 10^8 km^3 =9.05 \cdot 10^8 \cdot (10^9 m^3/k^3)=9.05\cdot 10^7 m^3$

So, the mass of the continent is

$m=\rho V = (2800)(9.05\cdot 10^{17})=2.5\cdot 10^{21} kg$

B)

The kinetic energy of a body is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the body

v is its speed

For the continent, we have:

$m=2.5\cdot 10^{21} kg$ is the mass

$v=4 cm/year$ is the speed

We have to convert the speed into SI units. we have:

1 cm = 0.01 m

$1 year = (365)(24)(60)(60) s = 3.15\cdot 10^7 s$

So, the speed is

$v=4 cm/year = 0.04 m/year \cdot \frac{1}{3.15\cdot 10^7}=1.27\cdot 10^{-9} m/s$

Therefore, the kinetic energy is

$K=\frac{1}{2}(2.5\cdot 10^{21} kg)(1.27\cdot 10^{-9} m/s)^2=2016 J$

C)

Again, the kinetic energy of an object is

$K=\frac{1}{2}mv^2$

For the jogger in this problem, his mass is

m = 80 kg

And we want its kinetic energy to be equal to that of the continent, so

K = 2016 J

Re-arranging the equation for v, we find what speed the jogger needs to have this kinetic energy:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2016)}{80}}=7.1 m/s$

Learn more about kinetic energy here:

brainly.com/question/6536722

#LearnwithBrainly

8 0

1 year ago

A cannon is mounted on a tower above a wide, level field. The barrel of the cannon is 20 m above the ground below. A cannonball

OLga [1]

<u>Answer:</u>

Cannonball will be in flight before it hits the ground for 2.02 seconds

<u>Explanation:</u>

Initial height from ground = 20 meter.

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate time when s = 20 meter.

Substituting

$20=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.02 seconds$

So it will take 2.02 seconds to reach ground.

5 0

1 year ago