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, 
⇒
Something that there is only so many of or not enough of
Answer:
10250 N/C leftwards
Explanation:
QA = 4 micro Coulomb
QB = - 5 micro Coulomb
AP = 6 m
BP = 2 m
A is origin, B is at 4 m and P is at 6 m .
The electric field due to charge QA at P is EA rightwards

The electric field due to charge QB at P is EB leftwards

The resultant electric field at P due the charges is given by
E = EB - EA
E = 11250 - 1000 = 10250 N/C leftwards
Answer:
Explanation:
1) The time of flight equation for projectile motion can be used here to find total time in air.
t = 2vsin∅ / g
where v is speed, Ф is launch angle
t = 2×4×sin 60 / 9.8
t = 0.71 seconds
2) Distance where it hit the ground is called as range and has the following standard equation
D = v² sin2Ф/g
D = 4²sin 2×60 / 9.8
D = 1.41m
3) Maximum elevation is maximum time reached
h = v² sin²Ф / 2g
h = 4²sin² 60 / 2*9.8
h = 0.61 m