What is one way to increase the amplitude if a wave in a medium?

One end of a steel rod of radius R = 10 mm and length L = 80cm is held in a vise. A force of magnitude F = 60kN is then applied

Alexandra [31]

Answer: 1.91*10^8 N/m²

Explanation:

Given

Radius of the steel, R = 10 mm = 0.01 m

Length of the steel, L = 80 cm = 0.8 m

Force applied on the steel, F = 60 kN

Stress on the rod, = ?

Area of the rod, A = πr²

A = 3.142 * 0.01²

A = 0.0003142

Stress = Force applied on the steel/Area of the steel

Stress = F/A

Stress = 60*10^3 / 0.0003142

Stress = 1.91*10^8 N/m²

From the calculations above, we can therefore say, the stress on the rod is 1.91*10^8 N/m²

8 0

2 years ago

A series RLC circuit has a resonant frequency = 6.00 kHz. When it is driven at a frequency = 8.00 kHz, it has an

ANEK [815]

The resistance (R) of the circuit is 707.1 ohms and the inductance (L) is 0.032 H.

<h3>Resistance of the circuit</h3>

For the phase constant of 45⁰, impedance is equal to the resistance of the circuit.

$Z= R\sqrt{2} \\\\R = \frac{Z}{\sqrt{2} } \\\\R = \frac{1000}{\sqrt{2} } = 707.1 \ ohms$

<h3>Resonant frequency</h3>

$f = \frac{1}{2\pi \sqrt{LC} } \\\\6000 = \frac{1}{2\pi \sqrt{LC} } \\\\2\pi(6000) = \frac{1}{\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi (6000)} \\\\LC = (\frac{1}{2\pi (6000)})^2\\\\LC = 7.034 \times 10^{-10} \\\\ C = \frac{7.034 \times 10^{-10} }{L} ---(1)$

<h3>At driven frequency</h3>

$X_l- X_c = R\\\\\omega L - \frac{1}{\omega C} = 707.1\\\\2\pi f L - \frac{1}{2\pi fC} = 707.1\\\\2\pi (8000) L - \frac{1}{2\pi (8000) C } = 707.1\ \ --(2)\\\\$