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dedylja [7]
2 years ago
14

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.
Physics
1 answer:
Alexus [3.1K]2 years 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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What conclusion can be derived by comparing the central tendencies of the two data sets?
zhannawk [14.2K]
The answer is B. I don’t think I need to explain this,
Mean is average, Mode is the most common number, and Median is the middle number when you put the numbers is numerical order from least to greatest
3 0
2 years ago
A 38 kg crate rests on a floor. A horizontal pulling force of 170 N is needed to start the crate
Mandarinka [93]

Answer:

0.456033049

Explanation:

F=\mu N where N=mg hence

F=\mu mg where m is mass of object, g is acceleration due to gravity whose value is taken as 9.81 m/s^{2}, \mu is the coefficient of static friction and F is the applied force.

Making \mu the subject we obtain

\mu=\frac {mg}{N} and substituting m for 38 Kg, g for 9.81 m/s^{2} and 170 N for  F we obtain

\mu=\frac{170} {38*9.81}=0.456033049

Therefore, the coefficient of static friction is 0.456033049

5 0
2 years ago
A 125g steel ball with a kinetic energy of .25j rolls along a horizontal track how high up an inclined track will the ball roll
Sauron [17]
E = 0.25 = m*g*h 
<span>h = 0.25/(m*g) = 0.25/(0.125*10) = 0.25/1.25 = 1/5 = 0.20 m
I hope this helps you have a great day and im sorry it took so long to get an answer</span>
6 0
2 years ago
g The international space station has an orbital period of 93 minutes at an altitude (above Earth's surface) of 410 km. A geosyn
krok68 [10]

Answer:

r = 4.21 10⁷ m

Explanation:

Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining

            T² = (\frac{4\pi }{G M_s} ) r³             (1)

           

in this case the period of the season is

            T₁ = 93 min (60 s / 1 min) = 5580 s

            r₁ = 410 + 6370 = 6780 km

            r₁ = 6.780 10⁶ m

for the satellite

           T₂ = 24 h (3600 s / 1h) = 86 400 s

if we substitute in equation 1

            T² = K r³

            K = T₁²/r₁³

            K = \frac{ 5580^2}{ (6.780 10^6)^2}

            K = 9.99 10⁻¹⁴ s² / m³

we can replace the satellite values

            r³ = T² / K

            r³ = 86400² / 9.99 10⁻¹⁴

            r = ∛(7.4724 10²²)

            r = 4.21 10⁷ m

this distance is from the center of the earth

7 0
2 years ago
As a rough approximation, the human body may be considered to be a cylinder of length L=2.0m and circumference C=0.8m. (To simpl
Brilliant_brown [7]

Answer:

Thermal Power = 460W

Explanation:

From Stephan-Boltzmann Law Formula;

P = єσT⁴A

Where,

P = Radiation energy

σ = Stefan-Boltzmann Constant

T = absolute temperature in Kelvin

є = Emissivity of the material.

A=Area of the emitting body

Now, σ = 5.67 x 10^(-8)

є = 0.6

Temperature = 30°C and coverting to kelvin = 30 + 273 = 303K

Area ; since we are to consider the sides of the human body as 2m and 0.8m,thus area = 2 x 0.8 = 1.6

Thus thermal power = 0.6 x 5.67 x 10^(-8) x303⁴ x 1.6 = 458. 8W

Normally, we approximate to the nearest 10W. Thus, thermal power is approximately 460W

4 0
2 years ago
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