The energy carried away in a one-quantum change is given by
where
h is the Planck constant
f is the frequency of vibration
For the system in this problem, the frequency of vibration is
, therefore the energy of one quantum of vibration is
Explanation:
Orbital period = 2 x 10⁶s
Total radius = 2.5 x 10¹²m
Unknown:
Tangential speed of the satellite = ?
Solution:
The tangential speed simply expresses the rate at which the distance moves around its orbit with time.
Since the orbit is taken to be circular;
Tangential speed =
r is the radius
t is the time or orbital radius
Tangential speed = }[/tex]
Tangential speed = 7.85 x 10⁶m/s
Learn more:
Circular motion brainly.com/question/2562955
#learnwithBrainly
Answer:
1.196 m
Explanation:
Given the wave equation :
y= 0.05 cos(5.25x-1775t)
Recall the general traverse wave relation :
y(x, t) = Acos(kx - wt)
A = Amplitude
To Obtian the wavelength ;
We compare the both equations :
Take the value of k ;
kx = 5.25x
k = 5.25
Recall;
k = 2π/λ
5.25 = 2π/λ
5.25λ = 2π
λ = 2π / 5.25
λ = (2 * 3.14) / 5.25 = 1.196 m
Answer:
Explanation:
Given,
mass of the bullet, m = 0.0233 Kg
Mass of the block, M = 2.41 Kg
horizontal spring constant, k = 845 N/m
Amplitude of oscillation, A = 0.196 m
Using conservation of energy when the bullet is embedded
PE = KE
Now using conservation of momentum to calculate the initial velocity of bullet