Answer:
a. A = 0.1656 m
b. % E = 1.219
Explanation:
Given
mB = 4.0 kg , mb = 50.0 g = 0.05 kg , u₁ = 150 m/s , k = 500 N / m
a.
To find the amplitude of the resulting SHM using conserver energy
ΔKe + ΔUg + ΔUs = 0
¹/₂ * m * v² - ¹/₂ * k * A² = 0
A = √ mB * vₓ² / k
vₓ = mb * u₁ / mb + mB
vₓ = 0.05 kg * 150 m / s / [0.050 + 4.0 ] kg = 1.8518
A = √ 4.0 kg * (1.852 m/s)² / (500 N / m)
A = 0.1656 m
b.
The percentage of kinetic energy
%E = Es / Ek
Es = ¹/₂ * k * A² = 500 N / m * 0.1656²m = 13.72 N*0.5
Ek = ¹/₂ * mb * v² = 0.05 kg * 150² m/s = 1125 N
% E = 13.72 / 1125 = 0.01219 *100
% E = 1.219
The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.
The relationship between the resistance and the resistivity of a wire is

where

is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length
the cross-sectional area is given by

where r is the radius of the wire. Substituting in the previous equation ,we find

For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is

Therefore, the new resistivity must be 4 times the original one.
Answer:
For left = 0 N/C
For right = 0 N/C
At middle =
N/C
Explanation:
Given data :-
б =
C/ m²
Considering the two thin metal plates to be non conducting sheets of charges.
Electric field is given by

1) To the left of the plate
= 0 N/C.
2) To the right of them.
= 0 N/C.
3) Between them.
=
=
=
N/C