Answer:
The maximum velocity is 1.58 m/s.
Explanation:
A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.
Spring constant, K = 100 N/m
mass, m = 0.1 kg
Amplitude, A = 5 cm = 0.05 m
Let the angular frequency is w.
The maximum velocity is
The vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.
The given parameters;
The components of the velocity can be horizontal or vertical velocity.
The vertical component of the velocity is affected by acceleration due to gravity while the horizontal component of the velocity is not affected by gravity.
The vertical component of the velocity is calculated as;
The horizontal component of the velocity is constant since it is not affected by gravity.
The horizontal component of the velocity = 16 m/s
Thus, the vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.
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Answer:
4
Explanation:
The kilogram-meter per second (kg · m/s or kg · m · s -1 ) is the standard unit of momentum . Reduced to base units in the International System of Units ( SI ), a kilogram-meter per second is the equivalent of a newton-second (N · s), which is the SI unit of impulse .
Answer:
J = 1800 kg-m/s
Explanation:
Given that,
Mass of a boy, m = 150 kg
Initial velocity of a boy, u = 12 m/s
Finally, it stops, v = 0
We need to find the impulse is required to produce this change in momentum. We know that impulse is equal to the change in momentum. So,
So, the impulse is equal to 1800 kg-m/s