Answer:
hehe
Explanation:
I dont know because I am a noob ant study
Answer:
La frecuencia será la misma en los dos medios, y en el vacio, no varia.
b) Vacío : n = 3,684 10 m 1,33 = 4 900 Å ,3161 10 m 55,1 n
,6 105 10 Hz
3,161 10 m
v 1,93 10 m/s 193548,38 km/s =
1,55
10 3 m/s
n
c
Vidrio v: =
,6 107 10 Hz
3,684 10 m
v 2,25 10 m/s 225 000km/s =
33,1
10 3 m/s
n
c
a
Answer:
Option (a)
Explanation:
We will discard options that don't fit the situation:
Option b: <em>Incorrect </em>since if the driver "hits the gas" then velocity is augmenting and it's not constant.
Option c and d: <em>Incorrect </em>since the situation doesn't give us any information that could be related directly to the terrain or movement direction.
Option a: Correct. At <em>stage 1</em> we can assume the driver was going at constant speed which means acceleration is constantly zero. At <em>stage 2 </em>we can assume the driver augmented speed linearly, this is, with constant positive acceleration. At <em>stage 3 </em>we can assume the driver slowed the speed linearly, with constant negative acceleration.
Wave speed = frequency (in Hz) x wavelength (in m) so wave speed = 11 x 0.011 = 0.121 m/s
These are the Kepler's laws of planetary motion.
This law relates a planet's orbital period and its average distance to the Sun. - Third law of Kepler.
The orbits of planets are ellipses with the Sun at one focus. - First law of Kepler.
The speed of a planet varies, such that a planet sweeps out an equal area in equal time frames. - Second law of Kepler.
