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.
- Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,
- So, to
![V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016](https://tex.z-dn.net/?f=V_x%3D%280.8%2B0.6%29c-%5B%5Cfrac%7B0.6c%2A%280.8c%29%5E2%7D%7Bc%5E2%7D%5D%3D1.016)
find 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,
- Substituting values, we get,
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:
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Answer:
Yeah it's right there from the one next to the exclamation point
When a neutron or a proton in the nucleus changes a gamma ray is produced (gamma rays are electromagnetic waves)
When an electron drops from a higher energy level to a lower energy level an electromagnetic wave is give off.
Answer:
If the frequency is doubled the wavelength is only half as long.If you put your fingertip in a pool of water and repeatedly move it up and down,you will create circular water waves.
Hope this helps!!!
If the primary wire's power is 10 A and one branch's power is 4 A, another branch's power will be 6A.
According to Kirchhoff's current law (KCL), the total current flowing through a parallel route circuit's junction equals the total current flowing away from it.
Provided that one of the two branches through which power exits the intersection has a flow of 4A, and also that the junction's overall flow entering it is 10A, the entire current going the junction should be 10A.
Consequently, the second wire's power may be expressed as;
I = I1+ I2 [ where I= total current (10A);
I1= current in one branch (4A) &
I2= current in another branch]
⇒I2 = I - I1
⇒I2 = 10A - 4A
⇒I2 = 6A
Therefore, it can be concluded that when the primary wire bears 10A power having 4A in one of its branches, another branch carries 6A power.
Learn more about Kirchhoff's law here:
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