Answer:
Explanation:
Let the plastic rod extends from - L to + L .
consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .
It will create a field at point P on y -axis . Distance of point P
= √ x² + .15²
electric field at P due to small charged length
dE = k λ dx x / (x² + .15² )
Its component along Y - axis
= dE cosθ where θ is angle between direction of field dE and y axis
= dE x .15 / √ x² + .15²
= k λ dx .15 / (x² + .15² )³/²
If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L
E = ∫ k λ .15 / (x² + .15² )³/² dx
= k λ x L / .15 √( L² / 4 + .15² )
Answer:
Explanation:
Gravity is a field force since the earth does not have to actually “touch” an object to pull it toward the earth. A magnetic force is a field force that attracts or repels another magnet. Likewise, electric charges cause attracting or repelling forces without actual contact between the charges
Answer:
Answer is a wave increasing in energy as it hits another wave.
Explanation:
I hope it's helpful!
An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7
In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.
As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.
Using the Newton second law;
![\sum F_x = \sum F_y = 0](https://tex.z-dn.net/?f=%5Csum%20F_x%20%3D%20%5Csum%20F_y%20%3D%200)
![\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }](https://tex.z-dn.net/?f=%5Cmathbf%7Bmg%20sin%20%5Ctheta%20-f_s%3D%20N-mgcos%20%5Ctheta%20%3D%200%20%7D)
![\mathbf{mg sin \theta =f_s}](https://tex.z-dn.net/?f=%5Cmathbf%7Bmg%20sin%20%5Ctheta%20%3Df_s%7D)
![\mathbf{mg sin \theta =\mu_s N}](https://tex.z-dn.net/?f=%5Cmathbf%7Bmg%20sin%20%5Ctheta%20%3D%5Cmu_s%20N%7D)
N = mg cos θ
Equating both force component together, we have:
![\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}](https://tex.z-dn.net/?f=%5Cmathbf%7Bmg%20sin%20%5Ctheta%20%3D%5Cmu_s%20%5C%20mg%20%5C%20cos%20%5Ctheta%7D)
![\mathbf{sin \theta =\mu_s \ \ cos \theta}](https://tex.z-dn.net/?f=%5Cmathbf%7Bsin%20%5Ctheta%20%3D%5Cmu_s%20%5C%20%5C%20cos%20%5Ctheta%7D)
![\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cmu_s%20%3D%20%5Cdfrac%7Bsin%20%5Ctheta%20%7D%7B%20cos%20%5Ctheta%7D%7D)
From trigonometry rule:
![\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}](https://tex.z-dn.net/?f=%5Cmathbf%7Btan%20%5Ctheta%3D%20%5Cdfrac%7Bsin%20%5Ctheta%20%7D%7B%20cos%20%5Ctheta%7D%7D)
∴
![\mathbf{\mu_s =\tan \theta}}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cmu_s%20%3D%5Ctan%20%5Ctheta%7D%7D)
![\mathbf{\mu_s =\tan 35^0}}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cmu_s%20%3D%5Ctan%2035%5E0%7D%7D)
![\mathbf{\mu_s = 0.700}}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cmu_s%20%3D%200.700%7D%7D)
Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7
Learn more about static friction here:
brainly.com/question/24882156?referrer=searchResults
Answer:
technically yes
Explanation:
with a gun depending on how fast it shoots so when you fire at something you shoot in front of it a little bit so you hit it but a plane that fast you shoot like 100 feet infront of it...