Answer:
The maximum amplitude (
) will be 7.96 V.
Explanation:
We know, for distortion free operation, the slew rate (S) of an OPAMP is written as

where '
' is the highest frequency signal.
Therefore, from the above equation we can write,

Answer:
L’énergie solaire récolte l’énergie radiante portée par la lumière de notre soleil en la convertissant en électricité.Biomasse des plantes. Les plantes sont capables d’exploiter et d’utiliser l’énergie lumineuse dans un processus appelé photosynthèse.
Answer:
It is explained in the explanation section
Explanation:
When the lift starts going downwards, it will start accelerating downwards. After a while, it will start moving with a constant velocity.
Constant velocity means that acceleration is zero and so the man will not feel any weight loss.
Now, Once the lift achieves constant velocity the acceleration is zero hence he will not experience any weight loss.
However, when the lift is in uniform motion, the lift and the man will fall down with an acceleration(a) that is less than that due to gravity(g) . Thus, the man will feel an apparent weight F which is not equal to zero.
Answer:
a) t=24s
b) number of oscillations= 11
Explanation:
In case of a damped simple harmonic oscillator the equation of motion is
m(d²x/dt²)+b(dx/dt)+kx=0
Therefore on solving the above differential equation we get,
x(t)=A₀
where A(t)=A₀
A₀ is the amplitude at t=0 and
is the angular frequency of damped SHM, which is given by,

Now coming to the problem,
Given: m=1.2 kg
k=9.8 N/m
b=210 g/s= 0.21 kg/s
A₀=13 cm
a) A(t)=A₀/8
⇒A₀
=A₀/8
⇒
applying logarithm on both sides
⇒
⇒
substituting the values

b) 

, where
is time period of damped SHM
⇒
let
be number of oscillations made
then, 
⇒
Take a look at a simple reaction like the one below:
In this reaction some reactant A is turned into some product B. The rate of reaction can be represented by a decrease in concentration of A over time or as the increase of B over time. This is written: