r₁ = distance of point A from charge q₁ = 0.13 m
r₂ = distance of point A from charge q₂ = 0.24 m
r₃ = distance of point A from charge q₃ = 0.13 m
Electric field by charge q₁ at A is given as
E₁ = k q₁ /r₁² = (9 x 10⁹) (2.30 x 10⁻¹²)/(0.13)² = 1.225 N/C towards right
Electric field by charge q₂ at A is given as
E₂ = k q₂ /r₂² = (9 x 10⁹) (4.50 x 10⁻¹²)/(0.24)² = 0.703 N/C towards left
Since the electric field in left direction is smaller, hence the electric field by the third charge must be in left direction
Electric field at A will be zero when
E₁ = E₂ + E₃
1.225 = 0.703 + E₃
E₃ = 0.522 N/C
Electric field by charge "q₃" is given as
E₃ = k q₃ /r₃²
0.522 = (9 x 10⁹) q₃/(0.13)²
q₃ = 0.980 x 10⁻¹² C = 0.980 pC
R1 + R4 = 1430 + 1350 = 2780 = R14 series combination of R1 & R4
R2 + R5 = 1350 + 1150 = 2500 = R25
The circuit has been reduced to 3 resistors in parallel
R314 = 2780 * 1100 / (2780 + 1100) = 788 this is the resistance of the parallel combination of R14 and R3
R31425 = 2500 * 788 / (2500 + 788) = 599 which is the equivalent of the circuit - you can also use the formula for 3 resistors in parallel but this seems simpler
Answer:
Speed of the helium after collision = 246 m/s
Explanation:
Given that
Mass of helium ,m₁ = 4 u
u₁=598 m/s
Mass of oxygen ,m₂ = 32 u
u₂ = 401 m/s
v₂ =445 m/s
Given that initially both are moving in the same direction and lets take they are moving in the right direction.
Speed of the helium after collision = v₁
There is no any external force on the masses that is why the linear momentum will be conserve.
Initial linear momentum = Final linear momentum
P = m v
m₁u₁+m₂u₂ = m₁v₁+m₂v₂
598 x 4 + 32 x 401 = 4 x v₁+ 32 x 445
v₁ = 246 m/s
Speed of the helium after collision = 246 m/s