Answer:
Explanation:
given,
op-amp circuit with a gain of = (Av₁) = 96 V/V
Band width = (Bw₁) = 8 kHz
Required bandwidth(Bw₂) = 32 kHz
Highest gain available =(Av₂) = ?
For the given system Bandwidth product is constant
Av₁ Bw₁ = Av₂ Bw₂
96 x 8 = Av₂ x 32
the highest gain available under these conditions
Answer:
M g H = 1/2 M v^2 potential energy = kinetic energy
v^2 = 2 g H = 2 * 9.80 * 6 = 117.6 m/s^2
v = 10.8 m/s
(C)
<h2>
Answer:</h2>
143μH
<h2>
Explanation:</h2>
The inductance (L) of a coil wire (e.g solenoid) is given by;
L = μ₀N²A / l --------------(i)
Where;
l = the length of the solenoid
A = cross-sectional area of the solenoid
N= number of turns of the solenoid
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
<em>From the question;</em>
N = 183 turns
l = 2.09cm = 0.0209m
diameter, d = 9.49mm = 0.00949m
<em>But;</em>
A = π d² / 4 [Take π = 3.142 and substitute d = 0.00949m]
A = 3.142 x 0.00949² / 4
A = 7.1 x 10⁻⁵m²
<em>Substitute these values into equation (i) as follows;</em>
L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209 [Take π = 3.142]
L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209
L = 143 x 10⁻⁶ H
L = 143 μH
Therefore the inductance in microhenrys of the Tarik's solenoid is 143
Answer:
no where is the main part of the question dude
