The y-component of the force on the particle at the given position is 120 N.
<h3>
Electric force on the particle</h3>
The electric force on the particle is determined by applying Coulomb's law and work-energy theorem as shown below;
Fd = W
Where;
- F is the applied force
- d is the distance
- W is potential
F = W/d
F = 60/0.5
F = 120 N
Thus, the y-component of the force on the particle at the given position is 120 N.
Learn more about electric force here: brainly.com/question/20880591
Answer:
When the string moves, it creates a very small change in the distance to the next point, th
Explanation:
When the string moves, it creates a very small change in the distance to the next point, this generates a restoring force that tends to push the string back, this small disturbance propagates along the string and is what creates the pulse.
This is similar to what happens when a spring is stretched and a restoring force is generated shaved by the law of shortening.
F = k Dx
Answer:
The ratio of kinetic energies of 5 kg object to 20 kg object is 1:1.
Explanation:
Kinetic energy is defined as energy possessed by an object due to its motion.It is calculated by:
![K.E=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=K.E%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
Kinetic energy of the 5 kg object.
Mass of object,m = 5 kg
Velocity of an object = v
![K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 5kg\times v^2](https://tex.z-dn.net/?f=K.E%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%205kg%5Ctimes%20v%5E2)
Kinetic energy of the 20 kg object.
Mass of object,m' = 20 kg
Velocity of an object = v'
![K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 20kg\times v'^2](https://tex.z-dn.net/?f=K.E%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%2020kg%5Ctimes%20v%27%5E2)
The ratio of the kinetic energy of the 5 kilogram object to the kinetic energy of the 20-kilogram object:
![\frac{K.E}{K.E'}=\frac{\frac{1}{2}\times 5kg\times v^2}{\frac{1}{2}\times 20kg\times v'^2}](https://tex.z-dn.net/?f=%5Cfrac%7BK.E%7D%7BK.E%27%7D%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%205kg%5Ctimes%20v%5E2%7D%7B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%2020kg%5Ctimes%20v%27%5E2%7D)
Given that, v = 2v'
![\frac{K.E}{K.E'}=\frac{1}{1}](https://tex.z-dn.net/?f=%5Cfrac%7BK.E%7D%7BK.E%27%7D%3D%5Cfrac%7B1%7D%7B1%7D)
The ratio of kinetic energies of 5 kg object to 20 kg object is 1:1.
Answer:
Lorentz force, the force exerted on a charged particle q moving with velocity v through an electric E and magnetic field B. The entire electromagnetic force F on the charged particle is called the Lorentz force (after the Dutch physicist Hendrik A. Lorentz) and is given by F = qE + qv × B.
Explanation:
N/A
Answer:
a.![2.86 m/s^2](https://tex.z-dn.net/?f=2.86%20m%2Fs%5E2)
b.1058 N
Explanation:
We are given that
Mass of each dog,M=18.5 kg
Mass of sled with rider,m=250 kg
a.Average force,F=185 N
![\mu_s=0.14](https://tex.z-dn.net/?f=%5Cmu_s%3D0.14)
![g=9.8 m/s^2](https://tex.z-dn.net/?f=g%3D9.8%20m%2Fs%5E2)
By Newton's second law
![8F-f=(8M+m)a](https://tex.z-dn.net/?f=8F-f%3D%288M%2Bm%29a)
![a=\frac{8F-f}{8M+m}=\frac{8(185)-(0.14)(9.8)(250)}{8(18.5)+250}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B8F-f%7D%7B8M%2Bm%7D%3D%5Cfrac%7B8%28185%29-%280.14%29%289.8%29%28250%29%7D%7B8%2818.5%29%2B250%7D)
![a=2.86 m/s^2](https://tex.z-dn.net/?f=a%3D2.86%20m%2Fs%5E2)
b.By Newton's second law
![T=ma+\mu_s mg](https://tex.z-dn.net/?f=T%3Dma%2B%5Cmu_s%20mg)
Substitute the values
![T=250\times 2.86+0.14(250)(9.8)=1058 N](https://tex.z-dn.net/?f=T%3D250%5Ctimes%202.86%2B0.14%28250%29%289.8%29%3D1058%20N)
Hence, the force in the coupling between the dogs and the sled=1058 N