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g100num [7]
2 years ago
10

A satellite orbiting Earth at a velocity of 3700 m/s collides with a piece of

Physics
1 answer:
Gekata [30.6K]2 years ago
7 0

Answer: B

Explanation:

You can use the conservation of momentum, under the assumption that no mass was lost when the collision occurred. The initial momentum of the system must equal the final momentum of the system. Our system is the region including, and only including, the satellite and the space debris. Classical momentum is defined as the product of mass and velocity:

p_i=p_f

m_1v_1_i+m_2v_2_i=m_1v_1_f+m_2v_2_f

Due to mass 1 equaling mass 2, we can factor these quantities out:

m(v_1_i+v_2_i)=m(v_1_f+v_2_f)

Cancel the mass term on both sides to get:

v_1_i+v_2_i=v_1_f+v_2_f

We have the initial and final velocities for everything besides the final velocity of the satellite. Plug everything in:

3700m/s+6000m/s=v_1_f+3700m/s

v_1_f=6000m/s

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2 years ago
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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