a = (152 - 80)/18 = 4 m/s2
d = v0t + ½at2 = 80*18 + ½*4*18^2 = 2088 m
Explanation:
It is given that,
Distance, r = 3.5 m
Electric field due to an infinite wall of charges, E = 125 N/C
We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :
It is clear that the electric field is inversely proportional to the distance. So,
E' = 291.67 N/C
So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.
Answer:
Find the set of value of x Find the set of value of x which satisfy the inequality 2r2- 5x 2 18which satisfy the inequality 2r2- Find the set of value of x which satisfy the inequality 2r2- 5x 2 185x 2 18
Answer:
F=19.79N
Explanation:
Force F=m×a, a being the acceleration.
a=dv/dt (speed variation/interval of time)
a=(0.640-0)/22ms=
0.640/0.022=640/22=29.1m/s²
Mass: m=680g=0.68kg
Force: F=0.68×29.1=19.79N