Explanation:
Given that,
Mass of the object, m = 7.11 kg
Spring constant of the spring, k = 61.6 N/m
Speed of the observer, 
We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :
Time period of oscillation measured by the observer is :
So, the time period of oscillation measured by the observer is 5.79 seconds.
Answer:
3.036×10⁻¹⁰ N
Explanation:
From newton's law of universal gravitation,
F = Gm1m2/r² .............................. Equation 1
Where F = Gravitational force between the balls, m1 = mass of the first ball, m2 = mass of the second ball, r = distance between their centers.
G = gravitational constant
Given: m1 = 7.9 kg, m2 = 6.1 kg, r = 2.0 m, G = 6.67×10⁻¹¹ Nm²/C²
Substituting into equation 1
F = 6.67×10⁻¹¹×7.9×6.1/2²
F = 321.427×10⁻¹¹/4
F = 30.36×10⁻¹¹
F = 3.036×10⁻¹⁰ N
Hence the force between the balls = 3.036×10⁻¹⁰ N
The answer is <span>C. 49 m/s
The kinetic equation is:
v2 = v1 + a * t
v1 - initial velocity
v2 - final velocity
a - gravitational acceleration
t - time
We know:
v2 = ?
v1 = 0 (in free fall
a = 9.8 m/s
t = 5
</span>v2 = v1 + a * t
v2 = 0 + 9.8 * 5
v2 = 0 + 49
v2 = 49 m/s
