360 degrees. I'm pretty sure
Answer: ![1.8(10)^{-4} rad](https://tex.z-dn.net/?f=1.8%2810%29%5E%7B-4%7D%20rad)
Explanation:
This problem is related to the Rayleigh Criterion, which provides the following formula to find the acuity or limit of resolution of an optic system with circular aperture (the eye in this case):
![\theta=1.22\frac{\lambda}{D}](https://tex.z-dn.net/?f=%5Ctheta%3D1.22%5Cfrac%7B%5Clambda%7D%7BD%7D)
Where:
is the angle of resolution (related to the acuity)
is the wavelength of the light
is the diameter of the pupil
Solving:
![\theta=1.22 \frac{554(10)^{-9} m}{3.75(10)^{-3} m}](https://tex.z-dn.net/?f=%5Ctheta%3D1.22%20%5Cfrac%7B554%2810%29%5E%7B-9%7D%20m%7D%7B3.75%2810%29%5E%7B-3%7D%20m%7D)
![\theta=1.8(10)^{-4} rad](https://tex.z-dn.net/?f=%5Ctheta%3D1.8%2810%29%5E%7B-4%7D%20rad)
Answer:
5.5 rad/s
Explanation:
The friction between the coin and the turntable provides the centripetal force that keeps the coin in circular motion. Therefore, we can write:
![\mu mg = m\omega^2 r](https://tex.z-dn.net/?f=%5Cmu%20mg%20%3D%20m%5Comega%5E2%20r)
where
is the coefficient of friction
m is the mass of the coin
is the acceleration of gravity
is the angular speed
r is the distance of the coin from the centre of rotation
In this problem,
r = 11.0 cm = 0.11 m
The coin starts to slip when the centripetal force becomes larger than the maximum frictional force:
![m\omega^2 r > \mu m g](https://tex.z-dn.net/?f=m%5Comega%5E2%20r%20%3E%20%5Cmu%20m%20g)
Solving for
, we find the angular speed at which this happens:
![\omega = \sqrt{\frac{\mu g}{r}}=\sqrt{\frac{(0.340)(9.8)}{0.11}}=5.5 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%5Cfrac%7B%5Cmu%20g%7D%7Br%7D%7D%3D%5Csqrt%7B%5Cfrac%7B%280.340%29%289.8%29%7D%7B0.11%7D%7D%3D5.5%20rad%2Fs)
Since two identical bulbs are connected in this circuit in series combination
So here same current will flow into the circuit
So here we can say
![V = V_1 + V_2](https://tex.z-dn.net/?f=V%20%3D%20V_1%20%2B%20V_2)
![V = 3 volts](https://tex.z-dn.net/?f=V%20%3D%203%20volts)
also we know that
![V_1 = iR_1](https://tex.z-dn.net/?f=V_1%20%3D%20iR_1)
![V_2 = iR_2](https://tex.z-dn.net/?f=V_2%20%3D%20iR_2)
also it is given that bulbs are identical so we have
![V_1 = V_2](https://tex.z-dn.net/?f=V_1%20%3D%20V_2)
as we have
![R_1 = R_2](https://tex.z-dn.net/?f=R_1%20%3D%20R_2)
so now we can say
![V = 2V_{CD}](https://tex.z-dn.net/?f=V%20%3D%202V_%7BCD%7D)
![3 = 2V_{CD}](https://tex.z-dn.net/?f=3%20%3D%202V_%7BCD%7D)
![V_{CD} = 1.5 Volts](https://tex.z-dn.net/?f=V_%7BCD%7D%20%3D%201.5%20Volts)
so here voltage across C and D is 1.5 volts