1. In a series circuit the sum of all voltage drops produced by the loads in the circuit

Harmonic excitation of motion is represent as

Gennadij [26K]

Harmonic excitation refers to a sinusoidal external force of a certain frequency applied to a system. ... Resonance occurs when the external excitation has the same frequency as the natural frequency of the system. It leads to large displacements and can cause a system to exceed its elastic range and fail structurally.

6 0

3 years ago

Match each context to the type of the law that is most suitable for it.

Bas_tet [7]

Answer:

sorry i dont understand the answer

Explanation:

but i think its a xd jk psml lol

5 0

3 years ago

Rough Riders Inc. manufactures jeans in the cutting and sewing process. Jeans are manufactured in 50-jean batch sizes. The cutti

miskamm [114]

Answer: me and ten points now

Explanation:

3 0

3 years ago

A 10MHz clock frequency is applied to a cascaded counter consisting of a modulus-5 counter, a modulus-8 counter and two modulus-

rewona [7]

If a clock frequency is applied to a cascaded counter, The lowest output frequency available will be

The lowest output frequency will be = $2.5kHz$

<h3>Cascade Counter</h3>

For a cascade counter,

Overall frequency = $5*8*10*10$

Overall frequency = $4000$

<h3>Lowest Frequency</h3>

Therefore,

the lowest frequency

$F = \frac{10*10^6}{4*10^3}\\\\F = 2500Hz\\\\F = 2.5kHz$

For more information on frequency, visit

brainly.com/question/17029587?referrer=searchResults

4 0

2 years ago

An ideal Otto cycle has a compression ratio of 9.2 and uses air as the working fluid. At the beginning of the compression proces

Allushta [10]

Answer:

(a) The amount of heat transferred to the air, $q_{out}$ is 215.5077 kJ/kg

(b) The net work output, $W_{net}$ , is 308.07 kJ/kg

(c) The thermal efficiency is 58.8%

(d) The Mean Effective Pressure, MEP, is 393.209 kPa

Explanation:

(a) The assumptions made are;

$c_p$ = 1.005 kJ/(kg·K), $c_v$ = 0.718 kJ/(kg·K), R = 0.287 kJ/(kg·K),

Process 1 to 2 is isentropic compression, therefore;

$T_{2}= T_{1}\left (\dfrac{v_{1}}{v_{2}} \right )^{k-1} = 300.15\times 9.2^{0.4} = 729.21 \, K$

From;

$\dfrac{p_{1}\times v_{1}}{T_{1}} = \dfrac{p_{2}\times v_{2}}{T_{2} }$

We have;

$p_{2} = \dfrac{p_{1}\times v_{1}\times T_{2}}{T_{1} \times v_{2}} = \dfrac{98\times 9.2\times 729.21}{300.15 } = 2190.43 \, kPa$

Process 2 to 3 is reversible constant volume heating, therefore;

$\dfrac{p_3}{T_3} =\dfrac{p_2}{T_2}$

p₃ = 2 × p₂ = 2 × 2190.43 = 4380.86 kPa

$T_3 = \dfrac{p_3 \times T_2}{p_2} =\dfrac{4380.86 \times 729.21}{2190.43} = 1458.42 \, K$

Process 3 to 4 is isentropic expansion, therefore;

$T_{3}= T_{4}\left (\dfrac{v_{4}}{v_{3}} \right )^{k-1}$

$1458.42= T_{4} \times \left (9.2 \right )^{0.4}$

$T_4 = \dfrac{1458.42}{(9.2)^{0.4}} = 600.3 \, K$

$q_{out} = m \times c_v \times (T_4 - T_1) = 0.718 \times (600.3 - 300.15) = 215.5077 \, kJ/kg$

The amount of heat transferred to the air, $q_{out}$ = 215.5077 kJ/kg

(b) The net work output, $W_{net}$ , is found as follows;

$W_{net} = q_{in} - q_{out}$

$q_{in} = m \times c_v \times (T_3 - T_2) = 0.718 \times (1458.42 - 729.21) = 523.574 \, kJ/kg$

$\therefore W_{net} = 523.574 - 215.5077 = 308.07 \, kJ/kg$

(c) The thermal efficiency is given by the relation;

$\eta_{th} = \dfrac{W_{net}}{q_{in}} \times 100= \dfrac{308.07}{523.574} \times 100= 58.8\%$

(d) From the general gas equation, we have;

$V_{1} = \dfrac{m\times R\times T_{1}}{p_{1}} = \dfrac{1\times 0.287\times 300.15}{98} =0.897\, m^{3}/kg$

The Mean Effective Pressure, MEP, is given as follows;

$MEP =\dfrac{W_{net}}{V_1 - V_2} = \dfrac{W_{net}}{V_1 \times (1- 1/r)}= \dfrac{308.07}{0.897\times (1- 1/9.2)} = 393.209 \, kPa$

The Mean Effective Pressure, MEP = 393.209 kPa.

3 0

3 years ago