Answer:
See attachments for step by step procedure into getting answers.
Explanation:
Given that;
R1 = 2.2 kΩ R2 = 4.7 kΩ C = 0.1 F Vdc = +5 V Vac = 5 V peak f = 1 kHz A. Use superposition to calculate the dc voltage at point X: Vdc = ______
R1 = 2.2 kΩ R2 = 4.7 kΩ C = 0.1 F Vdc = +5 V Vac = 5 V peak f = 1 kHz A. Use superposition to calculate the dc voltage at point X: Vdc = ______
R1 = 2.2 kΩ R2 = 4.7 kΩ C = 0.1 F Vdc = +5 V Vac = 5 V peak f = 1 kHz A. Use superposition to calculate the dc voltage at point X: Vdc = ______
See attacent for complete solving
Answer:
6.5 × 10¹⁵/ cm³
Explanation:
Thinking process:
The relation
With the expression Ef - Ei = 0.36 × 1.6 × 10⁻¹⁹
and ni = 1.5 × 10¹⁰
Temperature, T = 300 K
K = 1.38 × 10⁻²³
This generates N₀ = 1.654 × 10¹⁶ per cube
Now, there are 10¹⁶ per cubic centimeter
Hence,
Answer:
component of acceleration are a = 3.37 m/s² and ar = 22.74 m/s²
magnitude of acceleration is 22.98 m/s²
Explanation:
given data
velocity = 10 m/s
initial time to = 0
distance s = 400 m
time t = 14 s
to find out
components and magnitude of acceleration after the car has travelled 200 m
solution
first we find the radius of circular track that is
we know distance S = 2πR
400 = 2πR
R = 63.66 m
and tangential acceleration is
S = ut + 0.5 ×at²
here u is initial speed and t is time and S is distance
400 = 10 × 14 + 0.5 ×a (14)²
a = 3.37 m/s²
and here tangential acceleration is constant
so velocity at distance 200 m
v² - u² = 2 a S
v² = 10² + 2 ( 3.37) 200
v = 38.05 m/s
so radial acceleration at distance 200 m
ar =
ar =
ar = 22.74 m/s²
so magnitude of total acceleration is
A =
A =
A = 22.98 m/s²
so magnitude of acceleration is 22.98 m/s²