Answer:
Part a)
![KE = 77.95 J](https://tex.z-dn.net/?f=KE%20%3D%2077.95%20J)
Part b)
![L = 3.16 m](https://tex.z-dn.net/?f=L%20%3D%203.16%20m)
Part c)
distance L is independent of the mass of the sphere
Explanation:
Part a)
As we know that rotational kinetic energy of the sphere is given as
![KE = \frac{1}{2}I\omega_2 + \frac{1}{2}mv^2](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7DI%5Comega_2%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
so we will have
![KE = \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 + \frac{1}{2}mv^2](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7B2%7D%7B5%7DmR%5E2%29%28%5Cfrac%7Bv%7D%7BR%7D%29%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
so we will have
![KE = \frac{1}{5} mv^2 + \frac{1}{2}mv^2](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B5%7D%20mv%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![KE = \frac{7}{10} mv^2](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B7%7D%7B10%7D%20mv%5E2)
![KE = \frac{7}{10}(\frac{42}{9.81})(5.10^2)](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B7%7D%7B10%7D%28%5Cfrac%7B42%7D%7B9.81%7D%29%285.10%5E2%29)
![KE = 77.95 J](https://tex.z-dn.net/?f=KE%20%3D%2077.95%20J)
Part b)
By mechanical energy conservation law we know that
Work done against gravity = initial kinetic energy of the sphere
So we will have
![mgLsin\theta = KE](https://tex.z-dn.net/?f=mgLsin%5Ctheta%20%3D%20KE)
![\frac{42}{9.81}(9.81)L sin36 = 77.95](https://tex.z-dn.net/?f=%5Cfrac%7B42%7D%7B9.81%7D%289.81%29L%20sin36%20%3D%2077.95)
![L = 3.16 m](https://tex.z-dn.net/?f=L%20%3D%203.16%20m)
Part c)
by equation of energy conservation we know that
![\frac{7}{10}mv^2 = mgL sin\theta](https://tex.z-dn.net/?f=%5Cfrac%7B7%7D%7B10%7Dmv%5E2%20%3D%20mgL%20sin%5Ctheta)
so here we can see that distance L is independent of the mass of the sphere
Increased food production due to improved agricultural practices, control of many diseases by modern medicine and the use of energy to make historically uninhabitable areas of Earth inhabitable are examples of things which can extend carrying capacity.
In order for
sound to travel, it has to have its medium. The medium, which are particles or
molecules of solid, liquid or gas, collides with each other, it vibrates thus
creating sound. In space, there are no particles or molecules that will vibrate
for sound to be made.
Answer:
![\Delta p=1.3475\ kg-m/s](https://tex.z-dn.net/?f=%5CDelta%20p%3D1.3475%5C%20kg-m%2Fs)
Explanation:
The computation of magnitude of the change in the ball's momentum in kg · m/s is shown below:-
We represent
The ball mass = m = 275 g = 0.275 kg
Thus it goes to the floor and resurfaces upward.
The ball hits the ground at 3.30 m/s speed that is
u = -3.30 m/s which represents the Negative since the ball hits the ground)
It rebounds at a speed of 1.60 m / s i.e. v = 1.60 m/s (positive as the ball rebounds upstream)
![\Delta p=p_f-p_i](https://tex.z-dn.net/?f=%5CDelta%20p%3Dp_f-p_i)
![\Delta p=m(v-u)](https://tex.z-dn.net/?f=%5CDelta%20p%3Dm%28v-u%29)
![\Delta p=0.275\ kg(1.60\ m/s-(-3.30\ m/s))](https://tex.z-dn.net/?f=%5CDelta%20p%3D0.275%5C%20kg%281.60%5C%20m%2Fs-%28-3.30%5C%20m%2Fs%29%29)
![\Delta p=1.3475\ kg-m/s](https://tex.z-dn.net/?f=%5CDelta%20p%3D1.3475%5C%20kg-m%2Fs)
Given parameters:
Mass given = 1kg
Unknown:
Weight of the body at pole and equator = ?
Solution:
Both locations are on the surface of the earth. Generally, we take 9.8m/s² as the acceleration due to gravity on the earth.
Weight = mass x acceleration due to gravity
But little disparity occurs in the value of acceleration due to gravity from the pole to equator. This is due to equatorial bulge.
At the equator , 9.780 m/s²
pole 9.832 m/s²
Weight at equator = 1 x 9.780 = 9.78N
Weight at the pole = 1 x 9.832 = 9.832N