This force is centripetal force. it is the force that will hold the rider in the circular path and prevent him from skidding off. It is equal to the friction between the tyres and the road if no banking has been done on the road at the corner.
The electrostatic force between the two ions is ![2.9\cdot 10^{-10} N](https://tex.z-dn.net/?f=2.9%5Ccdot%2010%5E%7B-10%7D%20N)
Explanation:
The electrostatic force between two charged particle is given by Coulomb's law:
![F=k\frac{q_1 q_2}{r^2}](https://tex.z-dn.net/?f=F%3Dk%5Cfrac%7Bq_1%20q_2%7D%7Br%5E2%7D)
where
is the Coulomb's constant
are the two charges
r is the separation between the two charges
In this problem, the ion of sodium has a charge of
![q_1 = +e = +1.6\cdot 10^{-19} C](https://tex.z-dn.net/?f=q_1%20%3D%20%2Be%20%3D%20%2B1.6%5Ccdot%2010%5E%7B-19%7D%20C)
while the ion of chlorine has a charge of
![q_2 = -e = -1.6\cdot 10^{-19}C](https://tex.z-dn.net/?f=q_2%20%3D%20-e%20%3D%20-1.6%5Ccdot%2010%5E%7B-19%7DC)
And the distance between the two ions is
![r=282 pm = 282\cdot 10^{-12} m](https://tex.z-dn.net/?f=r%3D282%20pm%20%3D%20282%5Ccdot%2010%5E%7B-12%7D%20m)
Substituting, we find the electrostatic force between the two ions:
![F=(8.99\cdot 10^9) \frac{(1.6\cdot 10^{-19})(-1.6\cdot 10^{-19})}{(282\cdot 10^{-12})^2}=-2.9\cdot 10^{-10} N](https://tex.z-dn.net/?f=F%3D%288.99%5Ccdot%2010%5E9%29%20%5Cfrac%7B%281.6%5Ccdot%2010%5E%7B-19%7D%29%28-1.6%5Ccdot%2010%5E%7B-19%7D%29%7D%7B%28282%5Ccdot%2010%5E%7B-12%7D%29%5E2%7D%3D-2.9%5Ccdot%2010%5E%7B-10%7D%20N)
where the negative sign simply means that the force is attractive, since the two ions have opposite charge.
Learn more about electrostatic force:
brainly.com/question/8960054
brainly.com/question/4273177
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Answer:
Each carbon atom is covalently bonded to four other carbon atoms in diamond. A lot of energy is needed to separate the atoms. This is because covalent bonds are strong. This is the reason why diamond has a high melting point.
Explanation:
hey man here is your answer hope it helps
Answer:
a)![a=5.74\ g\ m/s^2](https://tex.z-dn.net/?f=a%3D5.74%5C%20g%5C%20m%2Fs%5E2)
b)![a=20.55\ g\ m/s^2](https://tex.z-dn.net/?f=a%3D20.55%5C%20g%5C%20m%2Fs%5E2)
Explanation:
Given that
Initial velocity u =0 m/s
Speed at 5 s v= 282 m/s
Rocket accelerated up to 5 s :
We know that
v= u + at
282= 0+ a x 5
![a=56.4\ m/s^2](https://tex.z-dn.net/?f=a%3D56.4%5C%20m%2Fs%5E2)
So the acceleration
![a=56.4\ m/s^2](https://tex.z-dn.net/?f=a%3D56.4%5C%20m%2Fs%5E2)
![a=5.74\ g\ m/s^2](https://tex.z-dn.net/?f=a%3D5.74%5C%20g%5C%20m%2Fs%5E2)
Rocket come to rest in 1.4 s :
Final velocity of rocket v=0 m/s
Initial velocity = 282 m/s
v= u + at
0= 282 - a x 1.4
![a=201.42\ m/s^2](https://tex.z-dn.net/?f=a%3D201.42%5C%20m%2Fs%5E2)
![a=20.55\ g\ m/s^2](https://tex.z-dn.net/?f=a%3D20.55%5C%20g%5C%20m%2Fs%5E2)
So the deceleration
![a=20.55\ g\ m/s^2](https://tex.z-dn.net/?f=a%3D20.55%5C%20g%5C%20m%2Fs%5E2)