Let’s use the *queue dramatic voice* LAW OF THE CONSERVATION OF MOMENTUM!
m1v1i + m2v2i = m1vf + m2vf
m1v1i - m1vf = m2vf - m2v2i
m1(v1i - vf) = m2(vf - v2i)
m2 = [m1(v1i - Vf)] / (vf - v2i)
m2 = [(1)(5 - -1)] / (-1 - -2)
m2 = 6 / 1
m2 = 6 kg
And that’s your final answer! Please press “Thanks”!
<span>Data:
Initial velocity upward: Vo = 5.00 m/s ,
Initial position: h = 40.0 m above the ground
Type of motion: free fall.
A) Compute the position of the sandbag at a time 1.05 s after its release.
Equation: y = h + Vo*t - g*(t^2) / 2
y = 40.0 m + 5.00 m/s * 1.05s - (9.8 m/s^2) * (1.05 s)^2 / 2 = 39.8 m
B)Compute the velocity of the sandbag at a time 1.05 s after its release.
Equation: Vf = Vo - g*t
=> Vf = 5.00 m/s - (9.8m/s^2) * (1.05 s) = - 5.29 m/s
Negative sign means that the sandbag is going down.
c) How many seconds after its release will the bag strike the ground?
Equation:
y = yo + Vo*t - g*(t^2) / 2
0 = 40.0 + 5.00t - 4.9 t^2
=> 4.9 t^2 - 5t - 40 = 0
Use the quadratic formula and you get: t = 3.41 s
</span>
I believe the answer is Natural Laws
The time taken by the pulse to travel from one support to the other is 0.208 s.
<h3>Given:</h3>
The mass of the cord is m = 0.65 kg.
The distance between the supports is, d = 8.0 m.
The tension in the cord is T = 120 N.
The time taken by the pulse to travel from one support to the other is given as,
Here, v is the linear velocity of a pulse. Its value is,
Then,
Thus, the time taken by the pulse to travel from one support to the other is 0.208 s.
Learn more about tension here:
brainly.com/question/24994188
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