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
An increase in a sound's pitch corresponds to an increase in the frequency!
Answer:
(4) A = 3 A, A₂ = 11 A
(5) 7 A
Explanation:
(4)
From the diagram,
A = 3+6+2
A = 11 A
V = A₂R
A₂ = V/R₂............ Equation 1
Given: V = 12 V, R₂ = 4 Ω
Substitute these values into equation 1
A₂ = 12/4
A₂ = 3 A
(5) Applying,
V = IR'
I = V/R'............ Equation 1
Where V = Voltage, I = cuurent, R' = total resistance.
But,
1/R' = (1/3)+(1/4)
1/R' = (3+4)/12
1/R' = 7/12
R' = 12/7 Ω
Given: V = 12 V
Substitute these values into equation 1
I = 12/(12/7)
I = 7 A
Therefore
A = 7 A
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