Answer:
a) the distance between her and the wall is 13 m
b) the period of her up-and-down motion is 6.5 s
Explanation:
Given the data in the question;
wavelength λ = 26 m
velocity v = 4.0 m/s
a) How far from the wall is she?
Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;
x = λ/2
we substitute
x = 26 m / 2
x = 13 m
Therefore, the distance between her and the wall is 13 m
b) What is the period of her up-and-down motion?
we know that the relationship between frequency, wavelength and wave speed is;
v = fλ
hence, f = v/λ
we also know that frequency is expressed as the reciprocal of the time period;
f = 1/T
Hence
1/T = v/λ
solve for T
Tv = λ
T = λ/v
we substitute
T = 26 m / 4 m/s
T = 6.5 s
Therefore, the period of her up-and-down motion is 6.5 s
T = 2 * pie √(L/g)
so, if length is increased by 9
then time period is increased by √9 = 3
hope it helped :)
True, the law of inertia effects both moving and non-moving objects.
Formula for final velocity: Vf= vi+(a*t)
Vi- initial velocity, a=acceleration, t-time
Vf=vi+(at)
Vf= 0+(9.8m/s*2.8s)
Vf= 27.44 m/s
The acceleration of the Earth when dropping something would be 9.8 m/s
Here is an reference that can help you answer problems like these.
Hope this helps and good luck :)