Constant force. Basicly in no gravity you have to keep puching smth for it to accelerate and once you stop it just would remain at the same speed. On earth the fact that something is falling faster and faster means it keeps beeing constantly pulled by gravity so the same principle applies
Answer with Explanation:
We are given that
Angle of incidence,![i=30^{\circ}](https://tex.z-dn.net/?f=i%3D30%5E%7B%5Ccirc%7D)
Angle of refraction,![r=19.24^{\circ}](https://tex.z-dn.net/?f=r%3D19.24%5E%7B%5Ccirc%7D)
a.Refractive index of air,![n_1=1](https://tex.z-dn.net/?f=n_1%3D1)
We know that
![n_2sinr=n_1sini](https://tex.z-dn.net/?f=n_2sinr%3Dn_1sini)
![n_2=\frac{n_1sin i}{sin r}=\frac{sin30}{sin19.24}=1.517](https://tex.z-dn.net/?f=n_2%3D%5Cfrac%7Bn_1sin%20i%7D%7Bsin%20r%7D%3D%5Cfrac%7Bsin30%7D%7Bsin19.24%7D%3D1.517)
b.Wavelength of red light in vacuum,![\lambda=632.8nm=632.8\times 10^{-9} m](https://tex.z-dn.net/?f=%5Clambda%3D632.8nm%3D632.8%5Ctimes%2010%5E%7B-9%7D%20m)
![1nm=10^{-9} m](https://tex.z-dn.net/?f=1nm%3D10%5E%7B-9%7D%20m)
Wavelength in the solution,![\lambda'=\frac{\lambda}{n_2}](https://tex.z-dn.net/?f=%5Clambda%27%3D%5Cfrac%7B%5Clambda%7D%7Bn_2%7D)
![\lambda'=\frac{632.8}{1.517}=417nm](https://tex.z-dn.net/?f=%5Clambda%27%3D%5Cfrac%7B632.8%7D%7B1.517%7D%3D417nm)
c.Frequency does not change .It remains same in vacuum and solution.
Frequency,![\nu=\frac{c}{\lamda}=\frac{3\times 10^8}{632.8\times 10^{-9}}](https://tex.z-dn.net/?f=%5Cnu%3D%5Cfrac%7Bc%7D%7B%5Clamda%7D%3D%5Cfrac%7B3%5Ctimes%2010%5E8%7D%7B632.8%5Ctimes%2010%5E%7B-9%7D%7D)
Where ![c=3\times 10^8 m/s](https://tex.z-dn.net/?f=c%3D3%5Ctimes%2010%5E8%20m%2Fs)
Frequency,![\nu=4.74\times 10^{14}Hz](https://tex.z-dn.net/?f=%5Cnu%3D4.74%5Ctimes%2010%5E%7B14%7DHz)
d.Speed in the solution,![v=\frac{c}{n_2}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bc%7D%7Bn_2%7D)
![v=\frac{3\times 10^8}{1.517}=1.98\times 10^8m/s](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B3%5Ctimes%2010%5E8%7D%7B1.517%7D%3D1.98%5Ctimes%2010%5E8m%2Fs)
A circle has a revolution of 360°. Since there are 12 hour markings, each hour interval has an angle of 30°. In radians, that would be equal to π/6 radians. So, in every 1 hour that passes, it covers π/6 of an angle. So, the angular velocity denoted as ω is π/6 ÷ 1 hour = π/6 rad/h. We can compute the average linear velocity, v, from the relationship:
v = rω, where r is the radius of the circle which is the length of the hour hand
v = (2.4 cm)(π/6 rad/h)
v = 1.257 cm/hour
Therefore, the average velocity is 1.257 cm per hour.
For the average acceleration, it is equal to zero. The hands of the clock move at a constant velocity. Since acceleration is the change of velocity per unit time, there is no change of velocity because it's constant. That's why it is zero.
Refer to the diagram shown below.
For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)
For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)
Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N
Answer:
T₂ = 339.06 N
T₃ = 276.57 N
Answer:
Simply,
<u>electrons</u> are "PARTICLES" orbiting the atoms, where, <u>current</u><u> </u>is the FLOW of some (free-to-move-around) electrons in a wire...