Self actualization.
hope this helps :3
The first one is: head
Second one is: 10 trillion km
Newton's third law of motion is naturally applied to collisions between two objects. In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction.For such a collision<span>, the forces acting between the two objects are equal in magnitude and opposite in direction</span>
<h2>
Option C is the correct answer.</h2>
Explanation:
Gravitational force is given by
![F=\frac{GMm}{r^2}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7BGMm%7D%7Br%5E2%7D)
G=6.674×10⁻¹¹ m³⋅kg⁻¹⋅s⁻²
M = Mass of object 1
m = mass of object 2
r = Distance between objects.
Here only variable is r value.
In case 1
![100=\frac{GMm}{(1000+500)^2}\\\\GMm=100\times 1500^2](https://tex.z-dn.net/?f=100%3D%5Cfrac%7BGMm%7D%7B%281000%2B500%29%5E2%7D%5C%5C%5C%5CGMm%3D100%5Ctimes%201500%5E2)
In case 2
![F=\frac{GMm}{(1000+500+500)^2}\\\\F=\frac{GMm}{2000^2}\\\\F=\frac{100\times 1500^2}{2000^2}\\\\F=56.25N](https://tex.z-dn.net/?f=F%3D%5Cfrac%7BGMm%7D%7B%281000%2B500%2B500%29%5E2%7D%5C%5C%5C%5CF%3D%5Cfrac%7BGMm%7D%7B2000%5E2%7D%5C%5C%5C%5CF%3D%5Cfrac%7B100%5Ctimes%201500%5E2%7D%7B2000%5E2%7D%5C%5C%5C%5CF%3D56.25N)
Option C is the correct answer.
Answer:
The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.
Explanation:
Given that,
Mass = 2 kg
Radius = 0.5 m
Angular speed = 3 rad/s
Force = 10 N
(I). We need to calculate the rotational kinetic energy
Using formula of kinetic energy
![K.E =\dfrac{1}{2}\timesI\omega^2](https://tex.z-dn.net/?f=K.E%20%3D%5Cdfrac%7B1%7D%7B2%7D%5CtimesI%5Comega%5E2)
![K.E=\dfrac{1}{2}\times mr^2\times\omega^2](https://tex.z-dn.net/?f=K.E%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20mr%5E2%5Ctimes%5Comega%5E2)
![K.E=\dfrac{1}{2}\times2\times(0.5)^2\times(3)^2](https://tex.z-dn.net/?f=K.E%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes2%5Ctimes%280.5%29%5E2%5Ctimes%283%29%5E2)
![K.E=2.25\ J](https://tex.z-dn.net/?f=K.E%3D2.25%5C%20J)
(II). We need to calculate the instantaneous change rate of the kinetic energy
Using formula of kinetic energy
![K.E=\dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=K.E%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
On differentiating
![\dfrac{K.E}{dt}=\dfrac{1}{2}m\times2v\times\dfrac{dv}{dt}](https://tex.z-dn.net/?f=%5Cdfrac%7BK.E%7D%7Bdt%7D%3D%5Cdfrac%7B1%7D%7B2%7Dm%5Ctimes2v%5Ctimes%5Cdfrac%7Bdv%7D%7Bdt%7D)
....(I)
Using newton's second law
![F = ma](https://tex.z-dn.net/?f=F%20%3D%20ma)
![a= \dfrac{F}{m}](https://tex.z-dn.net/?f=a%3D%20%5Cdfrac%7BF%7D%7Bm%7D)
![a=\dfrac{10}{2}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B10%7D%7B2%7D)
![a=5 m/s^2](https://tex.z-dn.net/?f=a%3D5%20m%2Fs%5E2)
Put the value of a in equation (I)
![\dfrac{K.E}{dt}=mva](https://tex.z-dn.net/?f=%5Cdfrac%7BK.E%7D%7Bdt%7D%3Dmva)
![\dfrac{K.E}{dt}=mr\omega a](https://tex.z-dn.net/?f=%5Cdfrac%7BK.E%7D%7Bdt%7D%3Dmr%5Comega%20a)
![\dfrac{K.E}{dt}=2\times0.5\times3\times5](https://tex.z-dn.net/?f=%5Cdfrac%7BK.E%7D%7Bdt%7D%3D2%5Ctimes0.5%5Ctimes3%5Ctimes5)
![\dfrac{K.E}{dt}=15\ J/s](https://tex.z-dn.net/?f=%5Cdfrac%7BK.E%7D%7Bdt%7D%3D15%5C%20J%2Fs)
Hence, The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.