Explanation:
At first it is in 14m position but position doesn't matter in displacement, similar case for time taken.
So at first it travels 6m in positive direction.
So displacement= 6m
Then it travels 13 in opposite or negative direction.
So displacement = 6 -13 m = -7 m
Hope it helps ya
PART a)
As we know that weight is product of mass and gravity
so here we have
![W = mg](https://tex.z-dn.net/?f=W%20%3D%20mg)
![W = 13(9.8) = 127.4 N](https://tex.z-dn.net/?f=W%20%3D%2013%289.8%29%20%3D%20127.4%20N)
Part B)
Normal force is counter balanced by weight of the crate
so here we have
N = W = 127.4 N
PART C)
As we know that
F = ma
now we have
m = 13 kg
F = 40 N
now we have
![40 = 13(a)](https://tex.z-dn.net/?f=40%20%3D%2013%28a%29)
![a = 3.08 m/s^2](https://tex.z-dn.net/?f=a%20%3D%203.08%20m%2Fs%5E2)
Part D)
Maximum static friction force is given as
![F_s = \mu_s N](https://tex.z-dn.net/?f=F_s%20%3D%20%5Cmu_s%20N)
here we know that
![\mu_s = 0.2](https://tex.z-dn.net/?f=%5Cmu_s%20%3D%200.2)
now we have
![F_s = 0.2(127.4) = 25.48 N](https://tex.z-dn.net/?f=F_s%20%3D%200.2%28127.4%29%20%3D%2025.48%20N)
PART E)
Since applied force on the block is
F = 12 N
Now since applied force is less than maximum static friction force
So it will not slide
So here friction force will be same as applied force
Static friction = 12 N
PART f)
Kinetic friction force is given as
![F_k = \mu_k N](https://tex.z-dn.net/?f=F_k%20%3D%20%5Cmu_k%20N)
here we know that
![\mu_k = 0.15](https://tex.z-dn.net/?f=%5Cmu_k%20%3D%200.15)
now friction force is
![F_k = 0.15(127.4) = 19.11 N](https://tex.z-dn.net/?f=F_k%20%3D%200.15%28127.4%29%20%3D%2019.11%20N)
now we have
![F_{net} = F - F_k](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20F%20-%20F_k)
![F_{net} = 40 - 19.11](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%2040%20-%2019.11)
![F_{net} = 20.89 N](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%2020.89%20N)
Part g)
as we know that
F = ma
![20.89 = 13a](https://tex.z-dn.net/?f=20.89%20%3D%2013a)
![a = 1.61 m/s^2](https://tex.z-dn.net/?f=a%20%3D%201.61%20m%2Fs%5E2)
The best answer is D. field lines should always be crossing each other.
Answer:
E ’= E / 8
therefore the correct answer is A
Explanation:
Let's calculate the electric field in an insulating sphere with a radius r <R, let's use Gauus's law, with a spherical Gaussian surface
Фi = ∫ E. dA =
/ε₀
E (4πr²) = q_{int} / ε₀
density is
ρ = q_{int} / V
q_{int} = ρ V = ρ 4/3 π r³
we substitute
E (4π r²) = ρ 4/3 π r³ /ε₀
E = 1 /3ε₀ ρ r
let's change the density by
ρ = Q / V = Q / (4/3 π R³)
E = 1 / 4πε₀ Q r / R³
if we now distribute the same charge on a sphere of radius R' = 2R
E ’= 1 / 4pieo Q r / (2R)³
E ’= 1 / 4ft Qr / R³ ⅛
E ’= E / 8
therefore the correct answer is A