Answer:
1.26 secs.
Explanation:
The following data were obtained from the question:
Force (F) = 20 N
Extention (e) = 0.2 m
Mass (m) = 4 Kg
Period (T) =.?
Next, we shall determine the spring constant, K for spring.
The spring constant, K can be obtained as follow:
Force (F) = 20 N
Extention (e) = 0.2 m
Spring constant (K) =..?
F = Ke
20 = K x 0.2
Divide both side by 0.2
K = 20/0.2
K = 100 N/m
Finally, we shall determine the period of oscillation of the 4 kg object suspended on the spring. This can be achieved as follow:
Mass (m) = 4 Kg
Spring constant (K) = 100 N/m
Period (T) =..?
T = 2π√(m/K)
T = 2π√(4/100)
T = 2π x √(0.04)
T = 2π x 0.2
T = 1.26 secs.
Therefore, the period of oscillation of the 4 kg object suspended on the spring is 1.26 secs.
Answer:
Her moment of inertia decreases causing her spin to speed up. The physics law behind this phenomenon is the conservation of angular momentum.
Explanation:
<em>Theory</em>
<u>The Law of conservation of angular momentum</u>
The angular momentum of a rotating body or a system remains constant unless it is acted upon by an external unbalance force.
Angular momentum = moment of inertia × angular velocity
The moment of inertia = mass×[perpendicular distance from axis of rotation]²
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⇒When skater draws in her outstretched arms the mass distribution get concentrated towards the axis of rotation so the moment of inertia of the body decrease.
But angular moment should conserve so angular velocity increases (spin increases)
Well,
The sun's energy is produced mainly in the core, which has sufficient temperature to initiate nuclear fusion.
Hydrogen --> Deuterium --> Tritium --> Helium --> Beryllium? --> Carbon --> ? --> Silicon --> Iron/Nickel = Most massive stars<span />
Answer:
A) The event horizon, singularity, and the chute located between the two.
Explanation:
Mass of the ball, m = 0.058 kg
Initial speed of the ball, u = 11 m/s
Final speed of the ball, v = -11 m/s (negative as it rebounds)
Time, t = 2.1 s
(a) Let F is the average force exerted on the wall. It is given by :
F = 0.607 N
(b) Area of wall, 
Let P is the average pressure on that area. It is given by :
P = 0.202 Pa
Hence, this is the required solution.
