If you go to high you’ll run out of oxygen and possibly be blown off due to high winds.
Well, if you're using the law to work with periods of Earth satellites,
then the most convenient unit is going to be 'hours' for the largest
orbits, or 'minutes' for the LEOs.
But if you're using it to work with periods of planets, asteroids, or
comets, then you'd be working in days or years.
Answer:
The vertical distance is ![d = \frac{2}{k} *[mg + f]](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B2%7D%7Bk%7D%20%2A%5Bmg%20%2B%20f%5D)
Explanation:
From the question we are told that
The mass of the cylinder is m
The kinetic frictional force is f
Generally from the work energy theorem
Here E the the energy of the spring which is increasing and this is mathematically represented as
Here k is the spring constant
P is the potential energy of the cylinder which is mathematically represented as
And
is the workdone by friction which is mathematically represented as
So
=> ![\frac{1}{2} * k * d^2 = d[mg + f ]](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20k%20%20%2A%20%20d%5E2%20%3D%20%20d%5Bmg%20%2B%20%20f%20%20%20%20%5D)
=> ![\frac{1}{2} * k * d = [mg + f ]](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20k%20%20%2A%20%20d%20%3D%20%20%5Bmg%20%2B%20%20f%20%20%20%20%5D)
=>
A gentle slope requires less force over a longer distance as compared to steep slope.
Explanation:
Mechanical advantage of a slope is equal to the ratio of length of slope and the height. A steep slope has shorter length as compared to a gentle slope for the same height. Therefore, mechanical advantage of a gentle slope is more than that of a steep slope. Hence, a gentle slope requires less force over a long distance than a steep slope.
It is False it's about how hard you work
