Higher frequency for lower wavelength...
.... If speed is fixed. ie no medium change.
Answer:
Explanation:
According to the <u>Third Kepler’s Law</u> of Planetary motion:
(1)
Where;:
is the period of the satellite
is the Gravitational Constant and its value is
is the mass of the Earth
is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).
On the other hand, the orbital velocity is given by:
(2)
Now, from (1) we can find
, in order to substitute this value in (2):
(3)
(4)
(5)
Substituting (5) in (2):
(6)
(7) This is the speed at which the satellite travels
Answer:
Time taken by Jesse to bring his bicycle =5.4 Sec
Explanation:
acceleration of bicycle = -2.5 m/sec²
Initial velocity of by cycle u =13.5 m/sec,
final velocity of bicycle v = 0 (since bicycle completely stop finally)
Let the time taken to stop the bicycle completely = t sec
We now by equation of motion,
putting the value of u,v & a in the equation
0=13.5 -2.5 x t ( a is taken as negative since velocity of bicycle is reducing)
= 5.4 Sec
Hertz, which is a unit for frequency (f), also means 1/second. Hence, the speed (S) of wave is the product of its wavelength (n) generated per second (frequency). The speed of the given wave above is,
S = nf ; S = (45 m) x (9 / s) = 405 m/s
Hence, the speed of the wave is 405 m/s.
I reckon it would be centimetres (cm).