express the following in metres (1) 52fm (2) 26 Mm (3)12am (4) 69 pm (5) 85 mm
1 answer:
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Answer:
a) Differential mode gain = 48
b) Common mode gain = 0.4
c) CMRR = 120
Explanation:
The output of a difference amplifier is related to the input by the equation:
When V₁ = 6.4 V, V₂ = 5.1 V and V₀ = 64.7 V, the equation becomes
6.4 A₁ + 5.1 A₂ = 64.7.....................(1)
When V₁ = 5.6 V, V₂ = 4.9 V and V₀ = 35.7 V, the equation becomes
5.6 A₁ + 4.9 A₂ = 35.7.....................(2)
Multiply equation (1) by 5.6 and (2) by 6.4
35.84 A₁ + 28.56A₂ = 362.32.....................(3)
35.84 A₁ + 31.36 A₂ = 228.48....................................(4)
Subtract equation (3) from (4)
2.8 A₂ = -133.84
A₂ = -133.84/2.8
A₂ = -47.8
Put the value of A₂ into equation (1)
6.4 A₁ + 5.1 (-47.8) = 64.7
6.4 A₁ = 64.7 + 243.78
A₁ = 308.48/6.4
A₁ = 48.2
a) Common mode gain = A₁ + A₂ = 48.2 + (-47.8)
Common mode gain = 0.4
b) Differential mode gain = (A₁ -A₂)/2
Differential mode gain = (48.2 - (-47.8))/2
Differential mode gain = 96/2
Differential mode gain = 48
c) Common Mode Rejection Ratio (CMRR)
(a) 198 g
When the rock is submerged into the water, there are two forces acting on the rock:
- its weight, equal to (m=mass, g=acceleration of gravity), downward
- the buoyant force, equal to (
=mass of water displaced), upward
So the resultant force, which is the apparent weight of the rock (W'), is
which can be rewritten as
where m' is the apparent mass of the rock. Using:
m = 540 g
m' = 342 g
we find the mass of water displaced
(b)
If the rock is completely submerged, the volume of the rock corresponds to the volume of water displaced.
The volume of water displaced is given by
where
is the mass of the water displaced
is the density of the water
Substituting,
And so this is also the volume of the rock.
(c)
The average density of the rock is given by
where
m = 540 g = 0.540 kg is the mass of the rock
is its volume
Substituting into the equation, we find