The answer is number two, number four, and number one
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
Answer:
v = 10 m/s
Explanation:
Let's assume the wheel does not slip as it accelerates.
Energy theory is more straightforward than kinematics in my opinion.
Work done on the wheel
W = Fd = 45(12) = 540 J
Some is converted to potential energy
PE = mgh = 4(9.8)12sin30 = 235.2 J
As there is no friction mentioned, the remainder is kinetic energy
KE = 540 - 235.2 = 304.8 J
KE = ½mv² + ½Iω²
ω = v/R
KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²
v = √(2KE / (m + I/R²))
v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6
v = 10.07968...
From the theory we know that:
c = λ / T
f = 1 / T
Where:
c = 3.
/ m (the speed of light)
λ is the wavelengh (in meters)
T is the period (in seconds)
f is the frequency (in Hz)
We were told that:
f = 7.30 . 
And we want to find out the value of λ.
c = λ / T
c = λ . 1/T
Swaping 1/T = f
c = λ . f
λ = c / f
λ = 3 . / 7.30 .
λ = 4.12 
m
Response: 4.12 m = 412 nm
:-)
