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
The product of (voltage) times (current, in Amperes) is POWER.
Answer:
the speed of the satellite is 12,880.53 km/h
Explanation:
Given;
radius of the circular orbit, r = 24,600 km
time taken to revolve around Earth, t = 12 hours
The circumference of the satellite is calculated as;
L = 2πr
L = 2π x 24,600 km
L = 49,200π km
L = 154,566.36 km
The speed of the satellite;
v = L/t
v = 154,566.36 / 12
v = 12,880.53 km/h
Therefore, the speed of the satellite is 12,880.53 km/h
Newtons Law of motion
HOPE IT HELPS:)
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