As per the question the charge of one coulomb is at 0 cm of the metre stick.the second charge of 4 coulomb is situated at at 100 cm of metre stick.
hence the separation distance between them is 100 cm.
now as per the question a proton is set up between them in such a way that the net force on it is zero
let the charge of proton is q coulomb let the proton is situated at distance of x cm from the charge 1 coulomb.hence it is situated at a distance of 100-x cm from the charge 4 coulomb.
the force exerted by 1 coulomb on proton is-
the force exerted by 4 coulomb on proton is-![\frac{1}{4\pi\epsilon} \frac{q*4}{[100-x]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon%7D%20%5Cfrac%7Bq%2A4%7D%7B%5B100-x%5D%5E2%7D)
as the net force is zero,hence-
![=\frac{1}{4\pi\epsilon} \frac{4*q}{[100-x]^2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon%7D%20%5Cfrac%7B4%2Aq%7D%7B%5B100-x%5D%5E2%7D)
![=\frac{1}{x^2} =\frac{4}{[100-x]^2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7Bx%5E2%7D%20%3D%5Cfrac%7B4%7D%7B%5B100-x%5D%5E2%7D)
![x^2=\frac{[100-x[^2}{4}[/tex[tex]x=\frac{100-x}{2}](https://tex.z-dn.net/?f=x%5E2%3D%5Cfrac%7B%5B100-x%5B%5E2%7D%7B4%7D%5B%2Ftex%3C%2Fp%3E%3Cp%3E%5Btex%5Dx%3D%5Cfrac%7B100-x%7D%7B2%7D)
cm [ans]
The velocity of the teddy bear as it strikes the ground is 7.67 m/s.
<h3>
Velocity of the teddy when it strikes the ground</h3>
The velocity of the teddy when it strikes the ground is calculated from principle of conservation of energy as shown below.
K.E(bottom) = P.E(top)
¹/₂mv² = mgh
v² = 2gh
v = √2gh
where;
- h is height of fall of the teddy
- g is acceleration due to gravity
v = √(2 x 9.8 x 3)
v = 7.67 m/s
Thus, the velocity of the teddy bear as it strikes the ground is 7.67 m/s.
Learn more about conservation of energy here: brainly.com/question/166559
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Answer:100m/s
Explanation:
in this question it does not specify the direction for a vector quantity .so in this question the velocity and the speed is considered the same.
Speed=d/t=200/2=100m/s
Speed=distance÷time so distance=speed×time therefore distance=4×2 which is 8.
