Answer:
i believe the answer is B only because i looked it up on g00gle so take what i say with a GRAIN of salt all puns intended
Explanation:
Answer:
A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation y(x,t)=(5.60cm)sin[(0.0340rad/cm)x]sin[(50.0rad/s)t]y(x,t)=(5.60cm)sin[(0.0340rad/cm)x]sin[(50.0rad/s)t], where the origin is at the left end of the string, the x-axis is along the string, and the y-axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation y(x, t) for this string if it were vibrating in its eighth harmonic?
Well, if your question is how light affects plants,
then you would want to design an experiment that plays aruond with the amount of light a plant gets
thus the thing changing (or variable) would be amount of light
Answer:
Option (2)
Explanation:
From the figure attached,
Horizontal component, 
![A_x=12[\text{Sin}(37)]](https://tex.z-dn.net/?f=A_x%3D12%5B%5Ctext%7BSin%7D%2837%29%5D)
= 7.22 m
Vertical component, ![A_y=A[\text{Cos}(37)]](https://tex.z-dn.net/?f=A_y%3DA%5B%5Ctext%7BCos%7D%2837%29%5D)
= 9.58 m
Similarly, Horizontal component of vector C,
= C[Cos(60)]
= 6[Cos(60)]
= 
= 3 m
![C_y=6[\text{Sin}(60)]](https://tex.z-dn.net/?f=C_y%3D6%5B%5Ctext%7BSin%7D%2860%29%5D)
= 5.20 m
Resultant Horizontal component of the vectors A + C,
m
= 4.38 m
Now magnitude of the resultant will be,
From ΔOBC,
= 
= 
= 6.1 m
Direction of the resultant will be towards vector A.
tan(∠COB) = 
= 
= 
m∠COB = 
= 46°
Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.
Option (2) will be the answer.
Explanation:
It is given that,
Frequency of monochromatic light, 
Separation between slits, 
(a) The condition for maxima is given by :
For third maxima,
(b) For second dark fringe, n = 2
Hence, this is the required solution.
