What happens if balanced forces are applied to a moving object? A. The object stops moving. B. The object moves faster in the sa
1 answer:
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Applying conservation of momentum;
Where m1 = 80.0 kg; v1 =4.10 m/s; m2 = 130.0 kg; v2 = 0; mf = (80+130) kg; vf = ??
Therefore,
80*4.1 + 130*0 = 210*vf
vf = (80*4.1)/210 = 1.56 m/s
Where m1 = 80.0 kg; v1 =4.10 m/s; m2 = 130.0 kg; v2 = 0; mf = (80+130) kg; vf = ??
Therefore,
80*4.1 + 130*0 = 210*vf
vf = (80*4.1)/210 = 1.56 m/s
Answer:
6 rad/s
Explanation:
In a spring the angular frequency is calculated as follows:
where is the angular frequency,
is the mass of the object in this case
, and
is the constant of the spring.
To calculate the angular frequency, first we need to find the constant which is calculated as follows:
Where is the force:
, and
is the distance from the equilibrium position:
.
Thus the spring constant:
And now we do have everything necessary to calculate the angular frequency:
the angular frequency of the oscillation is 6 rad/s