Rt= ΣR = 40Ω
Vt= 80V
It= 80V/40Ω= 2A
V1= 15Ω*2A= 30V
V2= 20Ω*2A= 40V
V3= 5Ω*2A= 10V
Answer:
The charge on the third object is − 21.7nC
Explanation:
From Gauss's Law
Φ = Q/ε₀
where;
Φ is the total electric flux through the shell = − 533 N⋅m²/C
Q is the total charge Q in the shell = ?
ε₀ is the permittivity of free space = 8.85 x 10⁻¹²
From this equation; Φ = Q/ε₀
Q = Φ * ε₀ = − 533 * 8.85 x 10⁻¹²
Q = −4.7 X 10⁻⁹ C = -4.7nC
Q = q₁ + q₂ + q₃
− 4.7nC = − 14.0 nC + 31.0 nC + q₃
− 4.7nC − 17nC = q₃
− 21.7nC = q₃
Therefore, the charge on the third object is − 21.7nC
Explanation:
It is given that,
Focal length of the concave mirror, f = -13.5 cm
Image distance, v = -37.5 cm (in front of mirror)
Let u is the object distance. It can be calculated using the mirror's formula as :



u = -21.09 cm
The magnification of the mirror is given by :


m = -1.77
So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.
Answer:
S = 2 π R
R (mean) = 92.9E6 miles
S = 2 * 3.14 * 92.9E6 miles = 5.84E8 miles