Answer:
uninstell this apps nobody will give u ans its happening to me alsoi was in exam hall i thought this app will give answer but no
Answer:
The answer is C (The cost of the solar panels can be very high.)
Answer:
ωf = 4.53 rad/s
Explanation:
By conservation of the angular momentum:
Ib*ωb = (Ib + Ic)*ωf
Where
Ib is the inertia of the ball
ωb is the initial angular velocity of the ball
Ic is the inertia of the catcher
ωf is the final angular velocity of the system
We need to calculate first Ib, Ic, ωb:
![Ib = mb*(L/2)^2=0.15*(1.2/2)^2=0.054 kg.m^2](https://tex.z-dn.net/?f=Ib%20%3D%20mb%2A%28L%2F2%29%5E2%3D0.15%2A%281.2%2F2%29%5E2%3D0.054%20kg.m%5E2)
![Ic = mc/12*L^2=2.2/12*1.2^2=0.264 kg.m^2](https://tex.z-dn.net/?f=Ic%20%3D%20mc%2F12%2AL%5E2%3D2.2%2F12%2A1.2%5E2%3D0.264%20kg.m%5E2)
ωb = Vb / (L/2) = 16 / (1.2/2) = 26.67 m/s
Now, ωf will be:
![\omega f = \frac{Ib*\omega b}{Ib + Ic} = 4.53rad/s](https://tex.z-dn.net/?f=%5Comega%20f%20%3D%20%5Cfrac%7BIb%2A%5Comega%20b%7D%7BIb%20%2B%20Ic%7D%20%20%3D%204.53rad%2Fs)
To solve this problem we will apply the concepts related to wavelength, as well as Rayleigh's Criterion or Optical resolution, the optical limit due to diffraction can be calculated empirically from the following relationship,
![sin\theta = 1.22\frac{\lambda}{d}](https://tex.z-dn.net/?f=sin%5Ctheta%20%3D%201.22%5Cfrac%7B%5Clambda%7D%7Bd%7D)
Here,
= Wavelength
d= Diameter of aperture
= Angular resolution or diffraction angle
Our values are given as,
![\theta = 11\°](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2011%5C%C2%B0)
The frequency of the sound is ![f = 9100 Hz](https://tex.z-dn.net/?f=f%20%3D%209100%20Hz)
The speed of the sound is ![v = 343 m/s](https://tex.z-dn.net/?f=v%20%3D%20343%20m%2Fs)
The wavelength of the sound is
![\lambda = \frac{v}{f}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7Bv%7D%7Bf%7D)
Here,
v = Velocity of the wave
f = Frequency
Replacing,
![\lambda = \frac{(343 m/s)}{(9100 Hz)}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7B%28343%20m%2Fs%29%7D%7B%289100%20Hz%29%7D)
![\lambda = 0.0377 m](https://tex.z-dn.net/?f=%5Clambda%20%3D%200.0377%20m)
The diffraction condition is then,
![sin\theta = 1.22\frac{\lambda}{d}](https://tex.z-dn.net/?f=sin%5Ctheta%20%3D%201.22%5Cfrac%7B%5Clambda%7D%7Bd%7D)
Replacing,
![sin(11\°) = 1.22\frac{(0.0377 m)}{(d)}](https://tex.z-dn.net/?f=sin%2811%5C%C2%B0%29%20%3D%201.22%5Cfrac%7B%280.0377%20m%29%7D%7B%28d%29%7D)
d = 0.24 m
Therefore the diameter should be 0.24m
Answer:
h = 78.4 m
Explanation:
Given that,
A stone takes 4s for the stone to touch the river's water surface.
We need to find the height of the bridge from the water level.
Let the height be h. We can use the second equation of motion to find h. So,
![h=ut+\dfrac{1}{2}gt^2\\\\ \because u=0\\\\So,\\\\h=\dfrac{1}{2}\times 9.8\times 4^2\\\\h=78.4\ m](https://tex.z-dn.net/?f=h%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dgt%5E2%5C%5C%5C%5C%20%5Cbecause%20u%3D0%5C%5C%5C%5CSo%2C%5C%5C%5C%5Ch%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%209.8%5Ctimes%204%5E2%5C%5C%5C%5Ch%3D78.4%5C%20m)
So, the height of the bridge from the water level is 78.4 m.