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Pavlova-9 [17]
3 years ago
7

Assume a small 11 kg crate is attached to a parachute. The chute pulls up on the crate with 150 N.

Physics
1 answer:
MaRussiya [10]3 years ago
3 0

Explanation:

a. The net force is the upward force of the chute minus the weight of the crate.

∑F = F − mg

∑F = 150 N − (11 kg) (9.8 m/s²)

∑F = 42.2 N

b. From Newton's second law, the net force equals the mass times acceleration:

∑F = ma

42.2 N = (11 kg) a

a = 3.84 m/s²

c. Acceleration is the change in velocity over change in time.  Assuming the crate is released from rest:

v = at + v₀

v = (3.84 m/s²) (5 s) + (0 m/s)

v = 19.2 m/s

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Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

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Substituting

h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9

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Here s = h,u = 600m/s a = -g and t = t

Substituting

h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\

Solving both equations

600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

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Passing of B occurs at 4108.31 height.

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\\ \sf\longmapsto ∆P=P

\\ \sf\longmapsto m_1v_1=m_2v_2

\\ \sf\longmapsto v_2=\dfrac{m_1v_1}{m_2}

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\\ \sf\longmapsto v_2=112.976/109

\\ \sf\longmapsto v_2\approx 1.3m/s

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