Answer:
this makes no since so i cant help you here sorry
Answer:
<h3>The answer is 40 N</h3>
Explanation:
The force acting on an object can be found by using the formula
<h3>force = mass × acceleration</h3>
From the question
mass = 4 kg
acceleration = 10 m/s²
So we have
force = 4 × 10
We have the final answer as
<h3>40 N</h3>
Hope this helps you
Answer:
Let I and j be the unit vector along x and y axis respectively.
Electric field at origin is given by
E= kq1/r1^2 i + kq2/r2^2j
= 9*10^9*1.6*10^-19*/10^-6*(2i+ j)
= (2.88i + 1.44j)*10^-3 N/C
Force on charge= qE= 3*10^-19*1.6*(2.88i +1. 44 j) *10^-3
F= (1.382 i + 0.691 j) *10^-21
Goodluck
Explanation:
Answer:
The Normal and Gravitational Force
Explanation:
The normal force pushes up and is between the ground and the scale. The gravitational force is the force exerted on the ground.
Complete question:
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?
Answer:
The electric field strength at the mid-point between the two rings is zero.
Explanation:
Given;
diameter of each ring, d = 10 cm = 0.1 m
distance between the rings, r = 21.0 cm = 0.21 m
charge of each ring, q = 40 nC = 40 x 10⁻⁹ C
let the midpoint between the two rings = x
The electric field strength at the midpoint between the two rings is given as;

Therefore, the electric field strength at the mid-point between the two rings is zero.