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Answer:
At the highest point the velocity is zero, the acceleration is directed downward.
Explanation:
This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.
I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.
At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.
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)