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spin [16.1K]
3 years ago
11

2. A Plate 0.02 mm distance from a fixed Plate moves at a velocity Of 0.6mls and requires a force of 1.962 N Per unit area to ma

itain this Speed. Determine the viscosity of the fluid between the plates?​
Physics
1 answer:
Marizza181 [45]3 years ago
5 0

Answer:

6.54 × 10⁻⁵ Pa-s

Explanation:

Since the shear force, F = μAu/y where μ = viscosity of fluid between plates, A = area of plates, u = velocity of fluid = 0.6 m/s and y = separation of plates = 0.02 mm = 2 × 10⁻⁵ m

Since F = μAu/y

F/A = μu/y where F/A = force per unit area

Since we are given force per unit area, F/A = 1.962 N per unit area = 1.962 N/m²

So,  μ = F/A ÷ u/y

substituting the values of the variables into the equation, we have

μ = F/A ÷ u/y

μ = 1.962 N/m² ÷ 0.6 m/s/2 × 10⁻⁵ m

μ = 1.962 N/m² ÷ 0.3 × 10⁵ /s

μ = 6.54 × 10⁻⁵ Ns/m²

μ = 6.54 × 10⁻⁵ Pa-s

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Oil at 150 C flows slowly through a long, thin-walled pipe of 30-mm inner diameter. The pipe is suspended in a room for which th
chubhunter [2.5K]

Answer:

<em>1.01 W/m</em>

Explanation:

diameter of the pipe d = 30 mm = 0.03 m

radius of the pipe r = d/2 = 0.015 m

external air temperature Ta = 20 °C

temperature of pipe wall Tw = 150 °C

convection coefficient at outer tube surface h = 11 W/m^2-K

From the above,<em> we assumed that the pipe wall and the oil are in thermal equilibrium</em>.

area of the pipe per unit length A = \pi r ^{2} = 7.069*10^{-4} m^2/m

convectional heat loss Q = Ah(Tw - Ta)

Q = 7.069 x 10^-4 x 11 x (150 - 20)

Q = 7.069 x 10^-4 x 11 x 130 = <em>1.01 W/m</em>

4 0
3 years ago
A parallel plate capacitor with plates of area A and plate separation d is charged so that the potential difference between its
faust18 [17]

Answer:

Explanation:

Given a parallel plate capacitor of

Area=A

Distance apart =d

Potential difference, =V

If the distance is reduce to d/2

What is p.d

We know that

Q=CV

Then,

V=Q/C

Then this shows that the voltage is inversely proportional to the capacitance

Therefore,

V∝1/C

So, VC=K

Now, the capacitance of a parallel plate capacitor is given as

C= εA/d

When the distance apart is d

Then,

C1=εA/d

When the distance is half d/2

C2= εA/(d/2)

C2= 2εA/d

Then, applying

VC=K

V1 is voltage of the full capacitor V1=V

V2 is the required voltage let say V'

Then,

V1C1=V2C2

V × εA/d=V' × 2εA/d

VεA/d = 2V'εA/d

Then the εA/d cancels on both sides and remains

V=2V'

Then, V'=V/2

The potential difference is half when the distance between the parallel plate capacitor was reduce to d/2

6 0
3 years ago
Read 2 more answers
The 2.5-Mg van is traveling with a speed of 100 km&gt;h when the brakes are applied and all four wheels lock. If the speed decre
Andrew [12]

Answer:

0.34

Explanation:

2.5 Mg = 2500 kg

The change in speed from 100 km/h to 40 km/h is

\Delta v = 100 - 40 = 60 km/h = 60 * 1000 / (60 * 60) = 16.67 m/s

The deceleration caused by friction force is the change in speed per unit of time

a = \Delta v / \Delta t = 16.67 / 5 = 3.33 m/s^2

Using Newton 2nd law we can calculate the friction force that caused this deceleration:

F = ma = 2500 * 3.33 = 8333 N

Let g = 9.8m/s2. Friction force is the product of normal (gravity) force and friction coefficient

F = mg\mu

8333 = 2500*9.8\mu

\mu = \frac{8333}{2500 * 9.8} = 0.34

4 0
3 years ago
What is the frequency of X-rays of 10 nm wavelength? 6 x 1016 Hz 3 x 1016 Hz 3.00 Hz 6 x 108 Hz
Evgesh-ka [11]

The frequency of any wave is  (wave speed) / (wavelength) .

The speed of light is  3 x 10⁸ m/s.
If the wavelength is 10 nm (10⁻⁸ m) , then the frequency is

                         (3 x 10⁸) / (10⁻⁸)  =  <em>3 x 10¹⁶ Hz</em>

That's 30 million gigahertz !

8 0
3 years ago
What is a disruption or change that is happening to the body that causes it to sweat ?
ivann1987 [24]

Answer:

the hypothalamus part of your brain sends a message to your body telling it to sweat.

When your internal temperature rises, your sweat glands release water to the surface of your skin.

4 0
3 years ago
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