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Kobotan [32]
3 years ago
8

3. If John leaves his house to walk his dog and he walks around the block, what is

Physics
1 answer:
satela [25.4K]3 years ago
8 0

Answer:

180 meters

Explanation:

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2. The second element is named as if they are treated like an anion but put in mind that these are no ions in a covalent compound but we put -ide on the second element as if it is an anion.

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A wave has a velocity of 24 m/s and a period of 3.0 s. Calculate the wavelength of the wave.
Katyanochek1 [597]

Velocity (unit:m/s) of the wave is given with the formula:

v=f∧,

where f is the frequency which tells us how many waves are passing a point per second (unit: Hz) and ∧ is the wavelength, which tells us the length of those waves in metres (unit:m)

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∧=v/f=24m/s/1/3=24*3=72m

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An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same directi
Arturiano [62]

The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

To find the answer, we have to know about the Lorentz transformation.

<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

                  V_x'=0.8c\\V=0.6c\\m=4*10^5kg

  • Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

                  KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}

  • So, to V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame.
  • We have an expression from Lorents transformation for relativistic law of addition of velocities as,

                      V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V

  • Substituting values, we get,

          V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c

  • Thus, the KE will be,

              KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J

Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

Learn more about frame of reference here:

brainly.com/question/20897534

SPJ4

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