Answer:
Explanation:
Total momentum of the system before the collision
.5 x 3 - 1.5 x 1.5 = -0.75 kg m/s towards the left
If v be the velocity of the stuck pucks
momentum after the collision = 2 v
Applying conservation of momentum
2 v = - .75
v = - .375 m /s
Let after the collision v be the velocity of .5 kg puck
total momentum after the collision
.5 v + 1.5 x .231 = .5v +.3465
Applying conservation of momentum law
.5 v +.3465 = - .75
v = - 2.193 m/s
2 ) To verify whether the collision is elastic or not , we verify whether the kinetic energy is conserved or not.
Kinetic energy before the collision
= 2.25 + 1.6875
=3.9375 J
kinetic energy after the collision
= .04 + 1.2 =1.24 J
So kinetic energy is not conserved . Hence collision is not elastic.
3 ) Change in the momentum of .5 kg
1.5 - (-1.0965 )
= 2.5965
Average force applied = change in momentum / time
= 2.5965 / 25 x 10⁻³
= 103.86 N
Well, if the bag is moving across the floor at a constant velocity. Then I would assume that the Fnet force is equal to 0, and thus the value of acceleration would also be equal to 0.
20 N north because He is pushing it 20 newtons north.
Explanation:
Formula which holds true for a leans with radii 
and 
and index refraction n is given as follows.
![\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%3D%20%28n%20-%201%29%20%5B%5Cfrac%7B1%7D%7BR_%7B1%7D%7D%20-%20%5Cfrac%7B1%7D%7BR_%7B2%7D%7D%5D)
Since, the lens is immersed in liquid with index of refraction 
. Therefore, focal length obeys the following.
![\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf_%7B1%7D%7D%20%3D%20%5Cfrac%7Bn%20-%20n_%7B1%7D%7D%7Bn_%7B1%7D%7D%20%5B%5Cfrac%7B1%7D%7BR_%7B1%7D%7D%20-%20%5Cfrac%7B1%7D%7BR_%7B2%7D%7D%5D)
![\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%28n%20-%201%29%7D%20%3D%20%5B%5Cfrac%7B1%7D%7BR_%7B1%7D%7D%20-%20%5Cfrac%7B1%7D%7BR_%7B2%7D%7D%5D)
and, 
or, 
= 32.4 cm
Using thin lens equation, we will find the focal length as follows.
Hence, image distance can be calculated as follows.
= 47.9 cm
Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.
According to the article "Nuclear shapes" by Renee Lucas the nucleus's shape is mainly modified by vibrational and rotational features happening within the cell. According to the article if i read correctly "near closed shells spherical shapes prevail, while between closed shells the large number of valence nucleons in orbit with large particle angular momentum leads to nuclei with large deformations leading them to not only maintain its shape but also alloying it to work.
