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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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Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted
galina1969 [7]

Answer:

1) When d_{o} < d_{i} (hence  d_{o} < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When d_{o} > d_{i} (and d_{o} < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}

Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}

Where:

f = Focal length of the mirror = R/2

d_{i} = Image distance from the mirror

d_{o} = Object distance from the mirror

h_{i} = Image height

h_{o} = Object height

d_{o} is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

d_{i} is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When d_{o} < d_{i} (hence  d_{o} < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When d_{o} > d_{i} (and d_{o} < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

5 0
3 years ago
ou see an image of yourself in a mirror because light waves are A) diffracted. B) reflected back at you. C) refracted through ai
Oksi-84 [34.3K]
The answer would be B) reflected back at you
6 0
3 years ago
Read 2 more answers
A parallel plate capacitor filled with air with plate area 2-cm2 and plate separation of 0.5 mm is connected to a 12 V batter an
pashok25 [27]

Answer:

a) The charge of the capacitor is 4.25x10⁻¹¹C

b) The charge of the capacitor is 4.25x10⁻¹¹C because the battery is disconnected.

c) The potential difference across the plates is 18 V

d) The work is 7.64x10⁻¹⁰J

Explanation:

The capacitance of the capacitor is equal to:

C=\frac{e_{0}A }{d}

A = 2 cm² = 0.0002 m²

d = 0.5 mm = 0.0005 m

Replacing:

C=\frac{8.85x10^{-12}*0.0002 }{0.0005} =3.54x10^{-12} F = 3.54pF

a) The charge of the capacitor is equal to:

Q = C*V = 3.54 * 12 = 42.48 pC = 4.25x10⁻¹¹C

b) The charge is the same because the battery is disconnected (Q = 4.25x10⁻¹¹C)

c) If distance is increased, we have:

C=\frac{8.85x10^{-12}*0.0002 }{0.00075} =2.36x10^{-12} F=2.36pF

The potential is:

V=\frac{Q}{C} =\frac{42.48}{2.36} =18V

d) The work done is equal to:

W=VQ=18*42.48=764.6pJ=7.64x10^{-10} J

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2 years ago
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What will be the gravitational force between two heavenly bodies if the masses of both are tripled keeping the distance between
Kruka [31]

Answer:

If the mass of one of the objects is tripled, then the force of gravity between them is tripled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces

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2 years ago
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pav-90 [236]
To answer this question, we will use the law of conservation of momentum which states that:
(m1+m2)Vi = m1V1 + m2V2 where:
m1 is the mass of the woman = 50 kg
m2 is the mass of the cart = 10 kg
Vi is the initial velocity (of woman and cart combined) = 5 m/sec
V1 is the final velocity of the woman = 7 m/sec
V2 is the final velocity of the cart that we need to calculate

Substitute with the givens in the above equation to get the final velocity of the cart as follows:
(50+10)(5) = (50)(7) + (10)V2
10V2 = -50
V2 = -5 m/sec
Note that the negative sign indicates that the cart is moving in an opposite direction to the others.
4 0
3 years ago
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