Answer:
The frequency of the oscillation is 2.45 Hz.
Explanation:
Given;
mass of the spring, m = 0.5 kg
total mechanical energy of the spring, E = 12 J
Determine the spring constant, k as follows;
E = ¹/₂kA²
kA² = 2E
k = (2E) / (A²)
k = (2 x 12) / (0.45²)
k = 118.519 N/m
Determine the angular frequency, ω;

Determine the frequency of the oscillation;
ω = 2πf
f = (ω) / (2π)
f = (15.396) / (2π)
f = 2.45 Hz
Therefore, the frequency of the oscillation is 2.45 Hz.
Answer:
answer is 11.76 meter
Explanation:
use 2nd equation of motion
S=ut+1/2at^2
The distance between Jupiter and the sun is 5.2 AU.
According to Kepler's third law, the square of the period of revolution of planets is proportional to the cube of their mean distances from the sun. From this; T^2 = r^3.
Now, we are told that the orbital period (T) is 11. 9 Earth years. We have to make the distance the subject of the formula.
r =T^2/3
r = (11.9)^2/3
r = 5.2 AU
Learn more: brainly.com/question/15207516
Answer:
The correct answer is:
(A) to the left
(B) at speed -0.8725 m/s
Explanation:
The given values are:
Plate 1:
Mass,
m₁ = 201 g
Velocity,
v₁ = +1.79 m/s
Plate 2:
Mass,
m₁ = 335 g
Velocity,
v₁ = -2.47 m/s
According to the conservation of momentum, we get
⇒ 
then,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒ 
⇒
(to the left)