object of mass m=2.6 kg is dropped from rest at a height of h above a massless spring with spring constant k=494.3 N/m that is i
1 answer:
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Answer:
Δθ₁ = 172.5 rev
Δθ₁h = 43.1 rev
Δθ₂ = 920 rev
Δθ₂h = 690 rev
Explanation:
- Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:
- Now, we need first to find the value of the angular acceleration, that we can get from the following expression:
- Since the machine starts from rest, ω₀ = 0.
- We know the value of ωf₁ (the operating speed) in rev/min.
- Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:
- Replacing by the givens in (2):
- Solving for α:
- Replacing (5) and Δt in (1), we get:
- in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:
- In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:
- First of all, we need to find the value of the angular acceleration during the second period.
- We can use again (2) replacing by the givens:
- ωf =0 (the machine finally comes to an stop)
- ω₀ = ωf₁ = 57.5 rev/sec
- Δt = 32 s
- Solving for α in (9), we get:
- Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:
- In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
Answer
given,
mass of the shell = 87 g = 0.087 Kg
speed of the muzzle = 853 m/s
mass of the helicopter = 4410 kg
A burst of 176 shell fired in 2.93 s
resulting average force = ?
momentum of the shell = m v
= 0.087 x 853
= 74.21 kgm/s
momentum of 176 shell is = 176 p
= 176 x 74.21
= 13060.96
momentum of helicopter = - 13060.96 kgm/s
amount of speed reduce a =
a=
a = 2.96 m/s²
velocity = \dfrac{2.96}{2.93}
v = 1.01 m/s