Answer:
a. E(r) = 3.67 x 10⁴ N/C
b. E(r) = 5.38 x 10⁴ N/C
Explanation:
r1 = 8.83 cm
r2 = 15.4 cm
Charge on the inner shell = 5.03 x 10^-⁸ C
Charge on the outer shell = 2.77 x 10^-⁸ C
a) To find the electric field at r = 11.1 cm,
r₁ = 8.83 cm < r = 11.1 cm < r₂ = 15.4 cm
E(r) = <u>1 </u> <u>q₁ </u>
4
∈₀ r₂
E(r) = <u>( 8.99 x 10⁹ N.m²/C² ) ( 5.03 x 10⁻⁸)</u>
(0.111m)²
E(r) = 3.67 x 10⁴ N/C
b) To find the electric field at r = 36.1 cm
since r₁ < r₂ < r = 36.1 cm
E(r) = <u>1 </u> <u>q₁ ₊ q₂ </u>
4∈₀ r₂
E(r) = <u>( 8.99 x 10⁹ N.m²/C² ) ( 5.03 + 2.77 )(1 x 10⁻⁸)</u>
(0.361m)²
E(r) = 5.38 x 10⁴ N/C
Answer:
Explanation:
To obtain the minimum acceleration, we know that normal force is equivalent to force in x-axis
Fx=N
#Frictional force is calculated using the formula:
but N= ma in this case acceleration in x axis hence
Answer:
The width of the slit is 0.4 mm (0.00040 m).
Explanation:
From the Young's interference expression, we have;
(λ ÷ d) = (Δy ÷ D)
where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.
Thus,
d = (Dλ) ÷ Δy
D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 ×
m)
d = (3.30 × 563 × ) ÷ (0.0047)
= 1.8579 × 
÷ 0.0047
= 0.0003951 m
d = 0.00040 m
The width of the slit is 0.4 mm (0.00040 m).
Answer:
anaemia, low blood pressure etc.
