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Anastaziya [24]
3 years ago
6

A plane drops a hamper of medical supplies from a height of 5000 m during a practice run over the ocean. The plane’s horizontal

velocity was 149 m/s at the instant the hamper was dropped. What is the magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s.
Physics
1 answer:
Marizza181 [45]3 years ago
4 0

Answer:

346.70015 m/s

Explanation:

In the x axis speed is

v_x=149\ m/s

In the y axis

v_y=\sqrt{2gh}\\\Rightarrow v_y=\sqrt{2\times 9.8\times 5000}

The resultant velocity is given by

v=\sqrt{v_x^2+v_y^2}\\\Rightarrow v=\sqrt{149^2+2\times 9.8\times 5000}\\\Rightarrow v=346.70015\ m/s

The magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean is 346.70015 m/s

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A wave pulse travels down a slinky. The mass of the slinky is m = 0.87 kg and is initially stretched to a length L = 6.8 m. The
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Answer:

1. v=14.2259\ m.s^{-1}

2. F_T=25.8924\ N

3. \lambda=29.6373\ m

Explanation:

Given:

  • mass of slinky, m=0.87\ kg
  • length of slinky, L=6.8\ m
  • amplitude of wave pulse, A=0.23\ m
  • time taken by the wave pulse to travel down the length, t=0.478\ s
  • frequency of wave pulse, f=0.48\ Hz=0.48\ s^{-1}

1.

\rm Speed\ of\ wave\ pulse=Length\ of\ slinky\div time\ taken\ by\ the\ wave\ to\ travel

v=\frac{6.8}{0.478}

v=14.2259\ m.s^{-1}

2.

<em>Now, we find the linear mass density of the slinky.</em>

\mu=\frac{m}{L}

\mu=\frac{0.87}{6.8}\ kg.m^{-1}

We have the relation involving the tension force as:

v=\sqrt{\frac{F_T}{\mu} }

14.2259=\sqrt{\frac{F_T}{\frac{0.87}{6.8}} }

202.3774=F_T\times \frac{6.8}{0.87}

F_T=25.8924\ N

3.

We have the relation for wavelength as:

\lambda=\frac{v}{f}

\lambda=\frac{14.2259}{0.48}

\lambda=29.6373\ m

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