Answer:
1.) 11 km/s
2.) 9.03 × 10^-5 metres
Explanation:
Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.
Electron q = 1.6×10^-19 C
Electron mass = 9.11×10^-31 Kg
(a) What is the speed of the electron 1.3 ns after entering this region?
E = F/q
F = Eq
Ma = Eq
M × V/t = Eq
Substitute all the parameters into the formula
9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19
V = 7.68×10^-18 /7.0×10^-22
V = 10971.43 m/s
V = 11 Km/s approximately
(b) How far does the electron travel during the 1.3 ns interval?
The initial velocity U = 64 km/s
S = ut + 1/2at^2
S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2
S =8.32×10^-5 + 7.13×10^-6
S = 9.03 × 10^-5 metres
IF voltage remains constant, then current is
inversely proportional to resistance.
The correct response is "b).", signifying "false" as the choice.
To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)
PART A) By definition the frequency in a spring is given by the equation

Where,
m = mass
k = spring constant
Our values are,
k=1700N/m
m=5.3 kg
Replacing,


PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

Where,
k = Spring constant
m = mass
y = Vertical compression
v = Velocity
This expression is equivalent to,

Our values are given as,
k=1700 N/m
V=1.70 m/s
y=0.045m
m=5.3 kg
Replacing we have,

Solving for A,



PART C) Finally, the total mechanical energy is given by the equation



Its the first one, second one, and last one.