Answer:![F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D%5Cfrac%7Bkq%5E2%7D%7B%28L%29%5E2%7D%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7D%2B%5Csqrt%7B2%7D%5Cright%20%5D)
Explanation:
Given
Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge
Force due to the charge placed at diagonally opposite end on -q charge
where 
Distance between the two charges
negative sign indicates that it is an attraction force
Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge
The magnitude of force by both the charge is same but at an angle of 
thus combination of two forces at 2 and 3 will be
Now it will add with force due to 1 charge
Thus net force will be
Answer:
t = 1.16 s.
Explanation:
Given,
speed of conveyor belt, v = 3.2 m/s
coefficient of friction,f = 0.28
Using newton second law
f = ma
and we also know that frictional force
f = μ N
f = μ m g
equating both the force equation
a = μ g
a = 0.28 x 9.81
a = 2.75 m/s²
Using Kinematic equation
v = u + at
3.2 = 0 + 2.75 x t
t = 1.16 s.
Time taken by the box to move without slipping is 1.16 s.
Answer:
Inductance, L = 0.0212 Henries
Explanation:
It is given that,
Number of turns, N = 17
Current through the coil, I = 4 A
The total flux enclosed by the one turn of the coil, 
The relation between the self inductance and the magnetic flux is given by :
L = 0.0212 Henries
So, the inductance of the coil is 0.0212 Henries. Hence, this is the required solution.
Answer:
4987N
Explanation:
Step 1:
Data obtained from the question include:
Mass (m) = 0.140 kg
Initial velocity (U) = 28.9 m/s
Time (t) = 1.85 ms = 1.85x10^-3s
Final velocity (V) = 37.0 m/s
Force (F) =?
Step 2:
Determination of the magnitude of the horizontal force applied. This can be obtained by applying the formula:
F = m(V + U) /t
F = 0.140(37+ 28.9) /1.85x10^-3
F = 9.226/1.85x10^-3
F = 4987N
Therefore, the magnitude of the horizontal force applied is 4987N
