Answer:
The recoil speed of Astronaut A is 0.26 m/s.
Explanation:
Given that,
Mass of astronaut A, ![m_A=110\ kg](https://tex.z-dn.net/?f=m_A%3D110%5C%20kg)
Mass of astronaut B, ![m_B=74\ kg](https://tex.z-dn.net/?f=m_B%3D74%5C%20kg)
Astronaut A pushes B away, with B attaining a final speed of 0.4, ![v_B=0.4\ m/s](https://tex.z-dn.net/?f=v_B%3D0.4%5C%20m%2Fs)
We need to find the recoil speed of astronaut A. The momentum remains conserved here. Using the law of conservation of linear momentum as :
![m_Av_A=m_Bv_B\\\\v_A=\dfrac{m_Bv_B}{m_A}\\\\v_A=\dfrac{74\times 0.4}{110}\\\\v_A=0.26\ m/s](https://tex.z-dn.net/?f=m_Av_A%3Dm_Bv_B%5C%5C%5C%5Cv_A%3D%5Cdfrac%7Bm_Bv_B%7D%7Bm_A%7D%5C%5C%5C%5Cv_A%3D%5Cdfrac%7B74%5Ctimes%200.4%7D%7B110%7D%5C%5C%5C%5Cv_A%3D0.26%5C%20m%2Fs)
So, the recoil speed of Astronaut A is 0.26 m/s.
(1,500 meters) x (1 sec / 330 meters) = (4 and 18/33) seconds
(4.55 sec, rounded)
Explanation:
We'll call the radius r and the diameter d:
We also assume that the riders are at a distance r = d/2 = 7m from the center of the wheel.
The period of the wheel is 24s. The tangent velocity of the wheel (and the riders) will be: (2pi/T)*r = 0.8 m/s (circa).
It means that in 3 minutes (180 seconds) they'll run 0.8 m/s * 180s = 144m.
Hopefully I understood the question. If yes, that's the answer.
Solution :
When the spacecraft is at halfway point, the distance from the Earth as well as Mars are same. We have to account the masses of the planets. The gravitational force that is exerted by the Earth is greater because of its combined mass with the space probe.
The mass of Earth is greater than the mass of Mars. Therefore, the force of Earth is more than Mars.