With what you do need help?????
Answer:
Part a)
charge on each sphere is -1.95 micro coulomb
Part b)
For first sphere
excess charge
For second sphere
excess charge
For third sphere
absent charge
For third sphere
absent charge
Explanation:
Part a)
Since all the spheres are of identical size so the total charge of the sphere will divide equally on them
So we have
So charge on each sphere is -1.95 micro coulomb
Part b)
For first sphere
initial charge = 2.2 micro coulomb
final charge = -1.95 micro coulomb
excess charge = -1.95 - 2.2 = -4.15 micro coulomb
Q = Ne
For second sphere
initial charge = 4.2 micro coulomb
final charge = -1.95 micro coulomb
excess charge = -1.95 - 4.2 = -6.15 micro coulomb
Q = Ne
For third sphere
initial charge = -7.4 micro coulomb
final charge = -1.95 micro coulomb
absent charge = -1.95 + 7.4 = 5.45 micro coulomb
Q = Ne
For third sphere
initial charge = -6.8 micro coulomb
final charge = -1.95 micro coulomb
absent charge = -1.95 + 6.8 = 4.85 micro coulomb
Q = Ne
Do u know newtons law of gravitation
F=Gm1*m2/r^2
F=attraction force =?
The area the radius vector will sweep is 0.889A
According to Kepler's second law, the radius vector <em>sweeps</em> out equal areas in equal times.
Let A = area and t = time period,
According to Kepler's law, A/t = constant
So, A₁/t₁ = A₂/t₂ where A₁ = area the radius vector sweeps at <em>slowest</em> orbital speed = A, t₁ = time period at <em>slowest </em>orbital speed = 45 days, A₂ = area the radius vector sweeps at<em> fastest</em> orbital speed, t₂ = time period at<em> fastest </em>orbital speed = 40 days.
Making A₂ subject of the formula, we have
A₂ = A₁t₂/t₁
Substituting the values of the variables into the equation, we have
A₂ = A₁t₂/t₁
A₂ = A × 40 days/45 days
A₂ = A × 40/45
A₂ = A × 8/9
A₂ = A × 0.889
A₂ = 0.889A
So, the area the radius vector will sweep is 0.889A
learn more about Kepler's second law here:
brainly.com/question/4639131