A ray of light crosses a boundary between two transparent materials. The medium the ray enters has a larger index of refraction.
1 answer:
Answer:
True, True, False, False, False, False.
Explanation:
The refraction index of a material is given by the formula n=c/v, where c is the speed of light in vacuum and v the speed of light in the material. If a ray of light crosses a boundary between two transparent materials and the medium the ray enters has a larger index of refraction it means that in this new medium the speed of light is smaller than on the other one, and then its wavelength is also reduced since f must remain the same (and ), otherwise there is a discontinuity on number of vibrations per second, which cannot happen. So we know that:
1) The wavelength of the light decreases as it enters into the medium with the greater index of refraction. True.
2) The frequency of the light remains constant as it transitions between materials. True.
3) The speed of the light remains constant as it transitions between materials. False.
4) The speed of the light increases as it enters the medium with the greater index of refraction. False.
5) The frequency of the light decreases as it enters into the medium with the greater index of refraction. False.
6) The wavelength of the light remains constant as it transitions between materials. False.
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Answer:
a) t=24s
b) number of oscillations= 11
Explanation:
In case of a damped simple harmonic oscillator the equation of motion is
m(d²x/dt²)+b(dx/dt)+kx=0
Therefore on solving the above differential equation we get,
x(t)=A₀
where A(t)=A₀
A₀ is the amplitude at t=0 and
is the angular frequency of damped SHM, which is given by,
Now coming to the problem,
Given: m=1.2 kg
k=9.8 N/m
b=210 g/s= 0.21 kg/s
A₀=13 cm
a) A(t)=A₀/8
⇒A₀ =A₀/8
⇒
applying logarithm on both sides
⇒
⇒
substituting the values
b)
, where
is time period of damped SHM
⇒
let be number of oscillations made
then,
⇒
A. The cliff was 30.7 m high
B. I also got 9.5 as the horizontal distance
Here is my work, I find making charts like this one to find knowns and unknowns can be helpful
B. I also got 9.5 as the horizontal distance
Here is my work, I find making charts like this one to find knowns and unknowns can be helpful
Answer
given,
position of particle
x(t)= A t + B t²
A = -3.5 m/s
B = 3.9 m/s²
t = 3 s
a) x(t)= -3.5 t + 3.9 t²
velocity of the particle is equal to the differentiation of position w.r.t. time.
------(1)
velocity of the particle at t = 3 s
v = -3.5 + 7.8 x 3
v = 19.9 m/s
b) velocity of the particle at origin
time at which particle is at origin
x(t)= -3.5 t + 3.9 t²
0 = t (-3.5 + 3.9 t )
t = 0,
t = 0 , 0.897 s
speed of the particle at t = 0.897 s
from equation (1)
v = -3.9 + 7.8 t
v = -3.9 + 7.8 x 0.897
v = 3.1 m/s