Answer:
Honestly for me it's a bit too blurry. Sorry luv.:(
Explanation:
Remember Coulomb's law: the magnitude of the electric force F between two stationary charges q₁ and q₂ over a distance r is

where k ≈ 8,98 × 10⁹ kg•m³/(s²•C²) is Coulomb's constant.
8.1. The diagram is simple, since only two forces are involved. The particle at Q₂ feels a force to the left due to the particle at Q₁ and a downward force due to the particle at Q₃.
8.2. First convert everything to base SI units:
0,02 µC = 0,02 × 10⁻⁶ C = 2 × 10⁻⁸ C
0,03 µC = 3 × 10⁻⁸ C
0,04 µC = 4 × 10⁻⁸ C
300 mm = 300 × 10⁻³ m = 0,3 m
600 mm = 0,6 m
Force due to Q₁ :

Force due to Q₃ :

8.3. The net force on the particle at Q₂ is the vector

Its magnitude is

and makes an angle θ with the positive horizontal axis (pointing to the right) such that

where we subtract 180° because
terminates in the third quadrant, but the inverse tangent function can only return angles between -90° and 90°. We use the fact that tan(x) has a period of 180° to get the angle that ends in the right quadrant.
Answer:
The final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .
Explanation:
Let us consider east as positive direction and west as negative direction .
Given
mass of puck 1 , 
mass of puck 2 , 
initial speed of puck 1 , 
initial speed of puck 2 , 
Final speed of puck 1 and puck 2 be
respectively
Apply conservation of linear momentum

=>
=>
-----(A)
Since collision is perfectly elastic , coefficient restitution e=1

=>
------(B)
From equation (A) and (B)

and 
Thus the final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .
Mass of the block = 1.4 kg
Weight of the block = mg = 1.4 × 9.8 = 13.72 N
Normal force from the surface (N) = 13.72 N
Acceleration = 1.25 m/s^2
Let the coefficient of kinetic friction be μ
Friction force = μN
F(net) = ma
μmg = ma
μg = a
μ = 
μ = 
μ = 0.1275
Hence, the coefficient of kinetic friction is: μ = 0.1275
Answer:
linear density of the string = 4.46 × 10⁻⁴ kg/m
Explanation:
given,
mass of the string = 31.2 g
length of string = 0.7 m
linear density of the string = 
linear density of the string = 
linear density of the string = 44.57 × 10⁻³ kg/m
linear density of the string = 4.46 × 10⁻⁴ kg/m