Answer:
Read below!
Explanation:
You can watch the sun wheel across the sky during the day, and the stars at night. Focus a telescope on any star besides the north star--especially southern stars--and you can watch them drift across your field of view.
An alternative explanation is that all the stars are painted on (or holes in) some canopy that rotates around the earth. This explanation does not account for the motion of the "wanderers," or planets, as the Greeks called them, or for the path of the moon among the stars.
As we know the stars are massive bodies of significant and varying distance to the earth, the notion they all swing around us in unison seems highly implausible
Answer:
![L = 2.746 cm](https://tex.z-dn.net/?f=L%20%3D%202.746%20cm)
Explanation:
As we know that thermal expansion coefficient of aluminium is given as
![\alpha = 24 \times 10^{-6} per ^oC](https://tex.z-dn.net/?f=%5Calpha%20%3D%2024%20%5Ctimes%2010%5E%7B-6%7D%20per%20%5EoC)
now we also know that after thermal expansion the final length is given as
![L = L_o(1 + \alpha \Delta T)](https://tex.z-dn.net/?f=L%20%3D%20L_o%281%20%2B%20%5Calpha%20%5CDelta%20T%29)
here we know that
![L_o = 2.739 cm](https://tex.z-dn.net/?f=L_o%20%3D%202.739%20cm)
![\alpha = 24 \times 10^{-6}](https://tex.z-dn.net/?f=%5Calpha%20%3D%2024%20%5Ctimes%2010%5E%7B-6%7D)
![\Delta T = 108 - 0= 108^oC](https://tex.z-dn.net/?f=%5CDelta%20T%20%3D%20108%20-%200%3D%20108%5EoC)
now we will have
![L = 2.739(1 + 24 \times 10^{-6} (108))](https://tex.z-dn.net/?f=L%20%3D%202.739%281%20%2B%2024%20%5Ctimes%2010%5E%7B-6%7D%20%28108%29%29)
![L = 2.746 cm](https://tex.z-dn.net/?f=L%20%3D%202.746%20cm)
da answer is liquiddddddddd
Explanation:
The given data is as follows.
mass = 0.20 kg
displacement = 2.6 cm
Kinetic energy = 1.4 J
Spring potential energy = 2.2 J
Now, we will calculate the total energy present present as follows.
Total energy = Kinetic energy + spring potential energy
= 1.4 J + 2.2 J
= 3.6 Joules
As maximum kinetic energy of the object will be equal to the total energy.
So, K.E = Total energy
= 3.6 J
Also, we know that
K.E = ![\frac{1}{2}mv^{2}_{m}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E%7B2%7D_%7Bm%7D)
or, v = ![\sqrt{\frac{2K.E}{m}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B2K.E%7D%7Bm%7D%7D)
= ![\sqrt{2 \times 3.6 J}{0.2 kg}](https://tex.z-dn.net/?f=%5Csqrt%7B2%20%5Ctimes%203.6%20J%7D%7B0.2%20kg%7D)
= ![\sqrt{36}](https://tex.z-dn.net/?f=%5Csqrt%7B36%7D)
= 6 m/s
thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.