42.9°
Explanation:
Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:
![x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;](https://tex.z-dn.net/?f=x%3A%5C%3B%5C%3B%5C%3B%5C%3BF%20-%20mg%5Csin%7B%5Ctheta%7D%20%3D%200%5C%3B%5C%3B%5C%3B%5C%3B)
![\Rightarrow mg\sin{\theta} = F](https://tex.z-dn.net/?f=%5CRightarrow%20mg%5Csin%7B%5Ctheta%7D%20%3D%20F)
Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at
Solving for the angle, we get
![\sin{\theta} = \dfrac{F}{mg}](https://tex.z-dn.net/?f=%5Csin%7B%5Ctheta%7D%20%3D%20%5Cdfrac%7BF%7D%7Bmg%7D)
or
![\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Csin%5E%7B-1%7D%5Cleft%28%5Cdfrac%7BF%7D%7Bmg%7D%5Cright%29)
![\;\;\;= \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]](https://tex.z-dn.net/?f=%5C%3B%5C%3B%5C%3B%3D%20%20%5Csin%5E%7B-1%7D%5Cleft%5B%5Cdfrac%7B34%5C%3A%5Ctext%7BN%7D%7D%7B%285.1%5C%3A%5Ctext%7Bkg%7D%29%289.8%5C%3A%5Ctext%7Bm%2Fs%7D%5E2%29%7D%5Cright%5D)
![\;\;\;=42.9°](https://tex.z-dn.net/?f=%5C%3B%5C%3B%5C%3B%3D42.9%C2%B0)
Answer:
9
Explanation:
2.13 rad/s * 26.9 sec
2.13 * 26.9
57.297
3282.88 deg / 360 deg = 9.12
It makes 9 complete revolutoins
Answer:
The width of the central bright fringe on the screen is observed to be unchanged is ![4.48*10^{-6}m](https://tex.z-dn.net/?f=4.48%2A10%5E%7B-6%7Dm)
Explanation:
To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes. Dark fringes in the diffraction pattern of a single slit are found at angles θ for which
![w sin\theta = m\lambda](https://tex.z-dn.net/?f=w%20sin%5Ctheta%20%3D%20m%5Clambda)
Where,
w = width
wavelength
m is an integer, m = 1, 2, 3...
We here know that as
as w are constant, then
![\frac{w_1}{\lambda_1} = \frac{w_2}{\lambda_2}](https://tex.z-dn.net/?f=%5Cfrac%7Bw_1%7D%7B%5Clambda_1%7D%20%3D%20%5Cfrac%7Bw_2%7D%7B%5Clambda_2%7D)
We need to find
, then
![w_2 = \frac{w_1}{\lambda_1}\lambda_2](https://tex.z-dn.net/?f=w_2%20%3D%20%5Cfrac%7Bw_1%7D%7B%5Clambda_1%7D%5Clambda_2)
Replacing with our values:
![w_2 = \frac{4.4*10^{-6}}{487}496](https://tex.z-dn.net/?f=w_2%20%3D%20%5Cfrac%7B4.4%2A10%5E%7B-6%7D%7D%7B487%7D496)
![w_2 = 4.48*10^{-6}m](https://tex.z-dn.net/?f=w_2%20%3D%204.48%2A10%5E%7B-6%7Dm)
Therefore the width of the central bright fringe on the screen is observed to be unchanged is ![4.48*10^{-6}m](https://tex.z-dn.net/?f=4.48%2A10%5E%7B-6%7Dm)