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, 
⇒
Kepler's second law of planetary motion<span> describes the speed of a </span>planet<span> traveling in an elliptical orbit around the sun. It states that a line between the sun and the </span>planetsweeps equal areas in equal times. Thus, the speed of theplanet<span> increases as it nears the sun and decreases as it recedes from the sun.</span>
I believe that answer is nitrogen.
First, you find the velocity at each component. The general equation is:
a = (v2 - v1)/t
a,x = (v2,x - v1,x)/t
-0.105 = (v2,x - 8.57)/6.67
v2,x = 7.87 m/s
a,y = (v2,y - v1,y)/t
0.101 = (v2,y - -2.61)/6.67
v2,y = -1.94 m/s
To find the final speed, find the resultant velocity by taking the hypotenuse.
v^2 = (v2,x)^2 + (v2,y)^2
v^2 = (7.87)^2 + (-1.94)^2
v = 8.1 m/s
Answer: A device records the time it takes sound waves to travel from the surface to the ocean floor and back again. Sound waves travel through water at a known speed. Once scientists know the travel time of the wave, they can calculate the distance to the ocean floor.
Explanation:
