Answer:
The kinetic energy of the particle will be 12U₀
Explanation:
Given that,
A particle is launched from point B with an initial velocity and reaches point A having gained U₀ joules of kinetic energy.
Constant force = 12F
According to question,
The kinetic energy is
....(I)
Constant force = 12F
A resistive force field is now set up ,
Resistive force is given by,
![F_{r}=12F](https://tex.z-dn.net/?f=F_%7Br%7D%3D12F)
When the particle moves from point B to point A then,
We need to calculate the kinetic energy
Using formula for kinetic energy
![U=F_{r}x](https://tex.z-dn.net/?f=U%3DF_%7Br%7Dx)
Put the value of ![F_{r}](https://tex.z-dn.net/?f=F_%7Br%7D)
![U=12Fx](https://tex.z-dn.net/?f=U%3D12Fx)
Now, from equation (I)
![U=12U_{0}](https://tex.z-dn.net/?f=U%3D12U_%7B0%7D)
Hence, The kinetic energy of the particle will be 12U₀.
Explanation:
Given that,
(a) Work done by the electric field is 12 J on a 0.0001 C of charge. The electric potential is defined as the work done per unit charged particles. It is given by :
![V=\dfrac{W}{q}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7BW%7D%7Bq%7D)
![V=\dfrac{12}{0.0001}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B12%7D%7B0.0001%7D)
![V=12\times 10^4\ Volt](https://tex.z-dn.net/?f=V%3D12%5Ctimes%2010%5E4%5C%20Volt)
(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :
![V=\dfrac{W}{q}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7BW%7D%7Bq%7D)
![V=\dfrac{24}{0.0002}](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7B24%7D%7B0.0002%7D)
![V=12\times 10^4\ Volt](https://tex.z-dn.net/?f=V%3D12%5Ctimes%2010%5E4%5C%20Volt)
Therefore, this is the required solution.
The net force is 270 N
Explanation:
We can solve this problem by using Newton's second law, which states that the net force on an object is equal to the product between its mass and its acceleration:
![F=ma](https://tex.z-dn.net/?f=F%3Dma)
where
F is the force
m is the mass
a is the acceleration
In this problem, we have
m = 90.0 kg
![a=3.0 m/s^2](https://tex.z-dn.net/?f=a%3D3.0%20m%2Fs%5E2)
Substituting, we find the net force on the object:
![F=(90.0)(3.0)=270 N](https://tex.z-dn.net/?f=F%3D%2890.0%29%283.0%29%3D270%20N)
Learn more about Newton's second law:
brainly.com/question/3820012
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