Answer:
Equal to the magnitude of F⃗
Explanation:
By Newton's third Law of motion, which states that:
<em>When</em><em> </em><em>two</em><em> </em><em>bodies</em><em> </em><em>interact</em><em> </em><em>they</em><em> </em><em>apply</em><em> </em><em>forces</em><em> </em><em>to</em><em> </em><em>one</em><em> </em><em> </em><em>a</em><em>m</em><em>other</em><em>,</em><em> </em><em>which</em><em> </em><em>are</em><em> </em><em>equal</em><em> </em><em>in</em><em> </em><em>magnitude</em><em> </em><em>and</em><em> </em><em>opposite</em><em> </em><em>in</em><em> </em><em>direction</em><em>.</em>
If the horse applies a force F⃗ on the cart, the cart must exert an equal and opposite force on the horse.
That is, the <u>magnitude</u> of the force applied by the horse on the cart is EQUAL to the <u>magnitude</u> of the force applied by the cart on the horse, but OPPOSITE in <u>direction</u><u>.</u>
Answer:
The initial charges of the spheres were q₁=6.7712×10⁻⁶C and q₂=-4.4350×10⁻⁶C.
Explanation:
As the spheres attract each other, the charges of the spheres are opposite.
The atracction force is given by:
F=![-\frac{Kq_{1}q_{2}}{r^{2}}](https://tex.z-dn.net/?f=-%5Cfrac%7BKq_%7B1%7Dq_%7B2%7D%7D%7Br%5E%7B2%7D%7D)
where:
K: Coulomb constant
q₁: charge of sphere 1
q₂: charge of sphere 2
r: distance between both charges
The electrostatic atraction force is 0.108 N so:
0.108N=-8.99×10⁹
![\frac{q_{1}q_{2}}{(0.5m)^2}](https://tex.z-dn.net/?f=%5Cfrac%7Bq_%7B1%7Dq_%7B2%7D%7D%7B%280.5m%29%5E2%7D)
q₁·q₂=![-\frac{0.108N*(0.5m)^2}{8.99*10^9\frac{N*m^2}{C^2}}](https://tex.z-dn.net/?f=-%5Cfrac%7B0.108N%2A%280.5m%29%5E2%7D%7B8.99%2A10%5E9%5Cfrac%7BN%2Am%5E2%7D%7BC%5E2%7D%7D)
q₁·q₂=-3.003×10⁻¹² C²
When the wire is connected the charges are equally distributed as the spheres are identical. Hence, the final charge is of each sphere is ![\frac{q_1q_2}{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bq_1q_2%7D%7B2%7D)
The repel force is 0.360 N and it is given by:
F=![\frac{K(\frac{q_1+q_2}{2})(\frac{q_1+q_2}{2})}{r^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7BK%28%5Cfrac%7Bq_1%2Bq_2%7D%7B2%7D%29%28%5Cfrac%7Bq_1%2Bq_2%7D%7B2%7D%29%7D%7Br%5E%7B2%7D%7D)
F=![\frac{K(q_1+q_2)^2}{4r^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7BK%28q_1%2Bq_2%29%5E2%7D%7B4r%5E%7B2%7D%7D)
Then, we get a secong equation:
(q₁+q₂)²=![\frac{0.360N*4*(0.5m)^2}{8.99*10^9\frac{N*m^2}{C^2}}](https://tex.z-dn.net/?f=%5Cfrac%7B0.360N%2A4%2A%280.5m%29%5E2%7D%7B8.99%2A10%5E9%5Cfrac%7BN%2Am%5E2%7D%7BC%5E2%7D%7D)
(q₁+q₂)=√4.004×10⁻¹¹ C²
q₁+q₂=6.3277×10⁻⁶ C
We solve the equation system:
![\left \{ {{q_1=\frac{-3.003*10^{-12}C^2}{q_2} } \atop {q_1+q_2=6.3277*10^{-6}C}} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bq_1%3D%5Cfrac%7B-3.003%2A10%5E%7B-12%7DC%5E2%7D%7Bq_2%7D%20%7D%20%5Catop%20%7Bq_1%2Bq_2%3D6.3277%2A10%5E%7B-6%7DC%7D%7D%20%5Cright.)
We replace q₁ in the second equation:
![\frac{-3.003*10^{-12}C^2}{q_2} +q_2=6.3277*10^{-6}C](https://tex.z-dn.net/?f=%5Cfrac%7B-3.003%2A10%5E%7B-12%7DC%5E2%7D%7Bq_2%7D%20%2Bq_2%3D6.3277%2A10%5E%7B-6%7DC)
![-3.003*10^{-12}C^2+(q_2)^2=6.3277*10^{-6}C*q_2](https://tex.z-dn.net/?f=-3.003%2A10%5E%7B-12%7DC%5E2%2B%28q_2%29%5E2%3D6.3277%2A10%5E%7B-6%7DC%2Aq_2)
![(q_2)^2-6.3277*10^{-6}C*q_2-3.003*10^{-12}C^2=0](https://tex.z-dn.net/?f=%28q_2%29%5E2-6.3277%2A10%5E%7B-6%7DC%2Aq_2-3.003%2A10%5E%7B-12%7DC%5E2%3D0)
The solutions are:
q₁=6.7712×10⁻⁶C
q₂=-4.4350×10⁻⁶C
To be honest I’m not sure you might want to ask Newton as he’s an expert best of luck
Answer:
It should be shortened. hope this helps!:)
Answer:
Playing hockey, driving a car, and even simply taking a walk are all everyday examples of Newton's laws of motion.