Answer:
tell ke first ,what will happen in zero gravity?
Answer:
- 7.088 m/s²
Explanation:
As we know that,
★ Acceleration = Change in velocity/Time
→ a = (v - u)/t
Here,
- Initial velocity (u) = 27.7 m/s
- Final velocity (v) = 10.9 m/s
→ a = (10.9 m/s - 27.7 m/s)/2.37 s
→ a = -16.8/2.37 m/s²
→ <u>a</u><u> </u><u>=</u><u> </u><u>-</u><u>7</u><u>.</u><u>0</u><u>8</u><u>8</u><u> </u><u>m/s²</u> [Answer]
Negative sign denotes that the velocity is decreasing.
Answer:
The time taken is 
Explanation:
From the question we are told that
The length of steel the wire is 
The length of the copper wire is 
The diameter of the wire is 
The tension is 
The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

Where
is the time taken to transverse the steel wire which is mathematically represented as
![t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]](https://tex.z-dn.net/?f=t_s%20%20%3D%20l_1%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B%5Crho%20%2A%20%5Cpi%20%2A%20%20d%5E2%20%7D%7B%204%20%2A%20%20T%7D%20%7D%20%5D)
here
is the density of steel with a value 
So
![t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]](https://tex.z-dn.net/?f=t_s%20%20%3D%2031%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B8920%20%2A%203.142%2A%20%20%281%2A10%5E%7B-3%7D%29%5E2%20%7D%7B%204%20%2A%20%20122%7D%20%7D%20%5D)

And
is the time taken to transverse the copper wire which is mathematically represented as
![t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]](https://tex.z-dn.net/?f=t_c%20%20%3D%20l_2%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B%5Crho_c%20%2A%20%5Cpi%20%2A%20%20d%5E2%20%7D%7B%204%20%2A%20%20T%7D%20%7D%20%5D)
here
is the density of steel with a value 
So
![t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]](https://tex.z-dn.net/?f=t_c%20%20%3D%2017%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B7860%20%2A%203.142%2A%20%20%281%2A10%5E%7B-3%7D%29%5E2%20%7D%7B%204%20%2A%20%20122%7D%20%7D%20%5D)

So



Data:
F (force) = ? (Newton)
k (<span>Constant spring force) = 50 N/m
x (</span>Spring deformation) = 15 cm → 0.15 m
Formula:

Solving:



Data:
E (energy) = ? (joule)
k (Constant spring force) = 50 N/m
x (Spring deformation) = 15 cm → 0.15 m
Formula:

Solving:(Energy associated with this stretching)




When the comet is closest to the Sun,
it has its maximum kinetic energy
and minimum gravitational potential energy. When the comet is far away from the Sun, it has maximum gravitational potential energy and minimal kinetic energy. It's faster when it's close because the Sun's gravity is pulling the comet closer. The opposite for when it gets farther away