Closed
Explanation:
A physical system that is isolated from its environment is said to be a closed system.
In a closed system, energy can be exchange but matter cannot be exchanged.
- A closed system sightly permits the only certain types of exchange.
- These systems are usually used to control and limit the impact of the environment on a system.
- Experiments can be conducted on the premises of any the interactions.
Learn more:
Momentum of an isolated system brainly.com/question/7973509
#learnwithBrainly
For astronomical objects, the time period can be calculated using:
T² = (4π²a³)/GM
where T is time in Earth years, a is distance in Astronomical units, M is solar mass (1 for the sun)
Thus,
T² = a³
a = ∛(29.46²)
a = 0.67 AU
1 AU = 1.496 × 10⁸ Km
0.67 * 1.496 × 10⁸ Km
= 1.43 × 10⁹ Km
Answer:
Loss, ![\Delta E=-10.63\ J](https://tex.z-dn.net/?f=%5CDelta%20E%3D-10.63%5C%20J)
Explanation:
Given that,
Mass of particle 1, ![m_1=m =0.66\ kg](https://tex.z-dn.net/?f=m_1%3Dm%20%3D0.66%5C%20kg)
Mass of particle 2, ![m_2=7.4m =4.884\ kg](https://tex.z-dn.net/?f=m_2%3D7.4m%20%3D4.884%5C%20kg)
Speed of particle 1, ![v_1=2v_o=2\times 6=12\ m/s](https://tex.z-dn.net/?f=v_1%3D2v_o%3D2%5Ctimes%206%3D12%5C%20m%2Fs)
Speed of particle 2, ![v_2=v_o=6\ m/s](https://tex.z-dn.net/?f=v_2%3Dv_o%3D6%5C%20m%2Fs)
To find,
The magnitude of the loss in kinetic energy after the collision.
Solve,
Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.
Applying the conservation of momentum to find the speed of two particles after the collision.
![m_1v_1+m_2v_2=(m_1+m_2)V](https://tex.z-dn.net/?f=m_1v_1%2Bm_2v_2%3D%28m_1%2Bm_2%29V)
![V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7Bm_1v_1%2Bm_2v_2%7D%7B%28m_1%2Bm_2%29%7D)
![V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B0.66%5Ctimes%2012%2B4.884%5Ctimes%206%7D%7B%280.66%2B4.884%29%7D)
V = 6.71 m/s
Initial kinetic energy before the collision,
![K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)](https://tex.z-dn.net/?f=K_i%3D%5Cdfrac%7B1%7D%7B2%7D%28m_1v_1%5E2%2Bm_2v_2%5E2%29)
![K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)](https://tex.z-dn.net/?f=K_i%3D%5Cdfrac%7B1%7D%7B2%7D%280.66%5Ctimes%2012%5E2%2B4.884%5Ctimes%206%5E2%29)
![K_i=135.43\ J](https://tex.z-dn.net/?f=K_i%3D135.43%5C%20J)
Final kinetic energy after the collision,
![K_f=\dfrac{1}{2}(m_1+m_2)V^2](https://tex.z-dn.net/?f=K_f%3D%5Cdfrac%7B1%7D%7B2%7D%28m_1%2Bm_2%29V%5E2)
![K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2](https://tex.z-dn.net/?f=K_f%3D%5Cdfrac%7B1%7D%7B2%7D%280.66%2B4.884%29%5Ctimes%206.71%5E2)
![K_f=124.80\ J](https://tex.z-dn.net/?f=K_f%3D124.80%5C%20J)
Lost in kinetic energy,
![\Delta K=K_f-K_i](https://tex.z-dn.net/?f=%5CDelta%20K%3DK_f-K_i)
![\Delta K=124.80-135.43](https://tex.z-dn.net/?f=%5CDelta%20K%3D124.80-135.43)
![\Delta E=-10.63\ J](https://tex.z-dn.net/?f=%5CDelta%20E%3D-10.63%5C%20J)
Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.
Answer:
I was also going to ask same question edited:ok i found its true