The strong nuclear force holds the nucleus of an atom together.
Somehow, it overcomes the electrical force of repulsion between protons in the nucleus, which all have the same charge but still stay close together somehow. (b)
Answer:
![\boxed{\sf Kinetic \ energy \ of \ the \ bear (KE) = 23002.1 \ J}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%20Kinetic%20%5C%20energy%20%5C%20of%20%5C%20the%20%5C%20bear%20%28KE%29%20%3D%2023002.1%20%5C%20J%7D%20)
Given:
Mass of the polar bear (m) = 6.8 kg
Speed of the polar bear (v) = 5.0 m/s
To Find:
Kinetic energy of the polar bear (KE)
Explanation:
Formula:
![\boxed{ \bold{\sf KE = \frac{1}{2} m {v}^{2} }}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Cbold%7B%5Csf%20KE%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20m%20%7Bv%7D%5E%7B2%7D%20%7D%7D)
Substituting values of m & v in the equation:
![\sf \implies KE = \frac{1}{2} \times 380.2 \times {11}^{2}](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20KE%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Ctimes%20380.2%20%5Ctimes%20%20%7B11%7D%5E%7B2%7D%20)
![\sf \implies KE = \frac{1}{ \cancel{2}} \times \cancel{2} \times 190.1 \times 121](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20KE%20%3D%20%5Cfrac%7B1%7D%7B%20%5Ccancel%7B2%7D%7D%20%20%5Ctimes%20%20%5Ccancel%7B2%7D%20%5Ctimes%20190.1%20%5Ctimes%20121%20)
![\sf \implies KE = 190.1 \times 121](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20KE%20%3D%20190.1%20%5Ctimes%20121%20)
![\sf \implies KE = 23002.1 \: J](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20KE%20%3D%2023002.1%20%5C%3A%20J)
![\therefore](https://tex.z-dn.net/?f=%20%5Ctherefore%20)
Kinetic energy of the polar bear (KE) = 23002.1 J
Find Acceleration
![\\ \rm\longmapsto F=ma](https://tex.z-dn.net/?f=%5C%5C%20%5Crm%5Clongmapsto%20F%3Dma)
![\\ \rm\longmapsto a=\dfrac{F}{m}](https://tex.z-dn.net/?f=%5C%5C%20%5Crm%5Clongmapsto%20a%3D%5Cdfrac%7BF%7D%7Bm%7D)
![\\ \rm\longmapsto a=\dfrac{-300}{60}](https://tex.z-dn.net/?f=%5C%5C%20%5Crm%5Clongmapsto%20a%3D%5Cdfrac%7B-300%7D%7B60%7D)
![\\ \rm\longmapsto a=-5m/s^2](https://tex.z-dn.net/?f=%5C%5C%20%5Crm%5Clongmapsto%20a%3D-5m%2Fs%5E2)
Now
Using 1st equation of kinematics
![\\ \rm\longmapsto v=u+at](https://tex.z-dn.net/?f=%5C%5C%20%5Crm%5Clongmapsto%20v%3Du%2Bat)
![\\ \rm\longmapsto t=\dfrac{v-u}{a}](https://tex.z-dn.net/?f=%5C%5C%20%5Crm%5Clongmapsto%20t%3D%5Cdfrac%7Bv-u%7D%7Ba%7D)
![\\ \rm\longmapsto t=\dfrac{20-50}{-5}](https://tex.z-dn.net/?f=%5C%5C%20%5Crm%5Clongmapsto%20t%3D%5Cdfrac%7B20-50%7D%7B-5%7D)
![\\ \rm\longmapsto t=\dfrac{-30}{-5}](https://tex.z-dn.net/?f=%5C%5C%20%5Crm%5Clongmapsto%20t%3D%5Cdfrac%7B-30%7D%7B-5%7D)
![\\ \rm\longmapsto t=6](https://tex.z-dn.net/?f=%5C%5C%20%5Crm%5Clongmapsto%20t%3D6)
Answer:
They weight the same, they're both 1 kilogram