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dybincka [34]
3 years ago
7

. A helicopter is 8.50 m above the ground and climbing at 2.5 m/s. It drops a package from rest (relative to the helicopter). At

the moment that the package is released, what is its velocity relative to the ground?
Physics
1 answer:
Rudik [331]3 years ago
4 0

Answer:

2.5 m/s

Explanation:

The velocity of the package relative to the ground = the velocity of the package relative to the helicopter + the velocity of the helicopter relative to the ground

v = 0 m/s + 2.5 m/s

v = 2.5 m/s

At the moment it is released, the package is rising at 2.5 m/s.

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An apparatus similar to the one used in lab uses an oscillating motor at one end to vibrate a long rope with frequency f = 40 Hz
Mice21 [21]

Answer:

The displacement in t = 0,  

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Explanation:

Given f = 40 Hz , A = 0.25m , μ = 0.02 kg / m, T = 20.48 N

v = √ T / μ

v = √20.48 N / 0.02 kg /m = 32 m/s

λ = v / f

λ = 32 m/s / 40 Hz = 0.8

K = 2 π / λ

K = 2π / 0.8 = 7.854

φ = X * 360 / λ

φ = 0.5 * 360 / 0.8 = 225 °

Using the model of y' displacement

y (t) = A* sin ( w * t - φ )

When t = 0

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5 0
3 years ago
A car initially traveling 7 m/s speeds up uniformly at a rate of 3 m/s2 until it reaches a velocity of 22 m/s. How much time did
dezoksy [38]

Answer:

t = 5 s

Explanation:

Data:

  • Initial Velocity (Vo) = 7 m/s
  • Acceleration (a) = 3 m/s²
  • Final Velocity (Vf) = 22 m/s
  • Time (t) = ?

Use formula:

  • \boxed{t=\frac{Vf - Vo}{a}}

Replace:

  • \boxed{t=\frac{22\frac{m}{s} -7\frac{m}{s}}{3\frac{m}{s^{2}}}}

Solve the subtraction of the numerator:

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It divides:

  • \boxed{t=5\ s}

How much time did it take the car to reach this final velocity?

It took a time of <u>5 seconds.</u>

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Answer:

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