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
Answer:
Step 1: List you assets
Step 2: List your liabilities
Step 3: Subtract your liabilities from your
assets
Step 4: Total your assets
Step 5: Total your liabilities
It makes no difference whether there are two bulbs or 1000 bulbs. If they are all in series then when one bulb burns out, or one little piece of wire between two bulbs comes loose, ALL of them go dark. In that situation it's a challenge to find where the break in the circuit is because all of the bulbs are dark.