from kinematics equation if we know that final speed is ZERO and initial speed is given that due to constant deceleration the object will stop in some distance "d" and this distance can be calculated by kinematics


here acceleration due to friction will be same at all different speed
so for 45 km/h speed the distance of stop is 15 m
while at other speed 112.5 km/h the distance will be unknown
now we will have


now divide above two equations


So it will stop in distance 93.75 m
Answer:
F = 2.26 × 10⁻³ N
Explanation:
given,
length of rod = 11 cm
charge = 19 nC
linear charge density = 3.9 x 10⁻⁷ C/m
electric force at 2 cm away.

F = E q

integrating from 0.02 to 0.02 + L
![F= \dfrac{2K\lambda\ q}{L}[ln(0.02+L)-ln(0.002)]](https://tex.z-dn.net/?f=F%3D%20%5Cdfrac%7B2K%5Clambda%5C%20q%7D%7BL%7D%5Bln%280.02%2BL%29-ln%280.002%29%5D)
![F= \dfrac{2\times 9 \times 10^9\times 3.9\times 10^{-7}\times 19 \times 10^{-9}}{0.11}[ln(0.02+0.11)-ln(0.002)]](https://tex.z-dn.net/?f=F%3D%20%5Cdfrac%7B2%5Ctimes%209%20%5Ctimes%2010%5E9%5Ctimes%203.9%5Ctimes%2010%5E%7B-7%7D%5Ctimes%2019%20%5Ctimes%2010%5E%7B-9%7D%7D%7B0.11%7D%5Bln%280.02%2B0.11%29-ln%280.002%29%5D)
F = 2.26 × 10⁻³ N
There is no illustration of the problem provided but I'll attempt to provide an answer.
The relationship between the electric potential difference between two points and the average strength of the electric field between those two points is given by:
║E║ = ΔV/d
║E║ is the magnitude of the average electric field, ΔV is the potential difference between A and B, and d is the distance between A and B.
We are given the following values:
║E║= 10N/C
d = 3m
Plug these values in and solve for ΔV
10 = ΔV/3
ΔV = 30V
Answer:
mass and distance
Explanation:
force is mass while motion can also be regard as distance or movement