Answer:
q₁ = + 1.25 nC
Explanation:
Theory of electrical forces
Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
Known data
q₃=5 nC
q₂=- 3 nC
d₁₃= 2 cm
d₂₃ = 4 cm
Graphic attached
The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.
For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).
The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).
The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)
Calculation of q1
F₁₃ = F₂₃
We divide by (k * q3) on both sides of the equation
q₁ = + 1.25 nC
Answer:
4.25 J
Explanation:
Given that
mass of plastic ball = 11 g
Mass of plastic ball = 0.011 kg
velocity of ball = 29 m/s
We know that from work power energy theorem
We know that kinetic energy of moving mass given as
Now by pitting the values
KE= 4.25 J
So the work done on the ball is 4.25 J
Answer:
the number of lines is 526
Explanation:
The wavelength λ =600nm = 600 × 10⁻⁶ mm
The diffraction angle θ = 39°
Recall the expression for the relation between the wavelength, angle and central maxima distance.
Recall the expression for the relation between the wavelength, angle and central maxima distance.
Recall the expression for the relation between the wavelength, angle and central maxima distance.
relation between the wave length, angle and central maxima distance
d = nλ / sinθ
Here n = 2 for second order maxima and d is the distance
= 2(600 × 10⁻⁶) / sin 39°
= 1200 × 10⁻⁶ / 0.6293
= 1.9 × 10⁻³ mm
N = 1/d
= 1 / 1.9 × 10⁻³
= 526
The grating has a line density of 526 lines per millimeter
Force is responsible for all changes in motion.
