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2 years ago

What are the three elementary parts of a vibrating system?

Engineering

1 answer:

zhenek [66]2 years ago

7 0

Answer:

the three part are mass, spring, damping

Explanation:

vibrating system consist of three elementary system namely

1) Mass - it is a rigid body due to which system experience vibration and kinetic energy due to vibration is directly proportional to velocity of the body.

2) Spring - the part that has elasticity and help to hold mass

3) Damping - this part considered to have zero mass and zero elasticity.

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An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the maximum air temperature and the rat

defon

Answer:

A) Rate of heat addition = 228.53 Hp

B) Maximum Air temperature = 1109 °C

Explanation:

We are given;

Cut off ratio; r_c = 1.5

Compression ratio; r = 18

Power produced; W` = 150 Hp

cp = 1.005 kJ/kg·K

cv = 0.718 kJ/kg·K

R = 0.287 kJ/kg·K

k = 1.4

Temperature;T1 = 17°C = 17 + 273 K = 290K

I've attached the rest of the explanation below.

8 0

2 years ago

Toeboards are usually ___ inches high and used on landings and balconies.

Ksju [112]

Answer:4 inches

Explanation:

Google told me

6 0

2 years ago

13–27. The conveyor belt is moving downward at 4 m>s. If the coefficient of static friction between the conveyor and the 15-k

Feliz [49]

Answer:

See explanation for step by step procedure to get answer.

Explanation:

Given that:

The conveyor belt is moving downward at 4 m>s. If the coefficient of static friction between the conveyor and the 15-kg package B is ms = 0.8, determine the shortest time the belt can stop so that the package does not slide on the belt.

See the attachments for complete steps to get answer.

4 0

2 years ago

A curve in a speed track has a radius of 1000 ft and a rated speed of 120 mi/h. (From Sample Prob. 12.7 is the definition of rat

forsale [732]

Answer:

$tan \theta = \frac{(176ft/s)^2}{1000 ft 32.2 ft/s^2}= 0.962$

$\theta = tan^{-1} (0.962) = 43.89$

Explanation:

If the question is: Determine the banking angle θ

We have the forces involved on the figure attached.

For this case we know that the weight is given by:

$W = mg$

And for this case the centripetal acceleration would be given by:

$a=\frac{v^2}{r}$

If we analyze the sum of forces on x and y we have:

$\sum F_x = m a_x$

$F + W sin \theta = ma cos theta$

And if we solve for the force we got:

$F = ma cos \theta - mg sin \theta = \frac{mv^2}{r} cos \theta - mg sin \theta$

$\sum F_y = m a_y$

$N - W cos \theta = ma sin \theta$

If we solve for the normal force we got:

$N =W cos \theta + ma sin \theta = \frac{mv^2}{r} sin \theta + mg cos \theta$

In order to find the banking angle we use the fact that F =0

$0 = \frac{mv^2}{r} cos \theta - mg sin \theta$

$tan \theta= \frac{v^2}{rg}$

The velocity on this case is 120 mi/h if we convert this into ft/ s we got:

$120 mi/h * \frac{5280 ft}{1mi} *\frac{1hr}{3600 s}= 176 ft/s$

And then we have this:

$tan \theta = \frac{(176ft/s)^2}{1000 ft 32.2 ft/s^2}= 0.962$

$\theta = tan^{-1} (0.962) = 43.89$

5 0

2 years ago

A three-phase line has an impedance of 0.4 j2.7 ohms per phase. The line feeds two balanced three-phase loads that are connected

Viktor [21]

Answer:

a. The magnitude of the line source voltage is

Vs = 4160 V

b. Total real and reactive power loss in the line is

Ploss = 12 kW

Qloss = j81 kvar

Sloss = 12 + j81 kVA

c. Real power and reactive power supplied at the sending end of the line

Ss = 540.046 + j476.95 kVA

Ps = 540.046 kW

Qs = j476.95 kvar

Explanation:

a. The magnitude of the line voltage at the source end of the line.

The voltage at the source end of the line is given by

Vs = Vload + (Total current×Zline)

Complex power of first load:

S₁ = 560.1 < cos⁻¹(0.707)

S₁ = 560.1 < 45° kVA

Complex power of second load:

S₂ = P₂×1 (unity power factor)

S₂ = 132×1

S₂ = 132 kVA

S₂ = 132 < cos⁻¹(1)

S₂ = 132 < 0° kVA

Total Complex power of load is

S = S₁ + S₂

S = 560.1 < 45° + 132 < 0°

S = 660 < 36.87° kVA

Total current is

I = S*/(3×Vload) ( * represents conjugate)

The phase voltage of load is

Vload = 3810.5/√3

Vload = 2200 V

I = 660 < -36.87°/(3×2200)

I = 100 < -36.87° A

The phase source voltage is

Vs = Vload + (Total current×Zline)

Vs = 2200 + (100 < -36.87°)×(0.4 + j2.7)

Vs = 2401.7 < 4.58° V

The magnitude of the line source voltage is

Vs = 2401.7×√3

Vs = 4160 V

b. Total real and reactive power loss in the line.

The 3-phase real power loss is given by

Ploss = 3×R×I²

Where R is the resistance of the line.

Ploss = 3×0.4×100²

Ploss = 12000 W

Ploss = 12 kW

The 3-phase reactive power loss is given by

Qloss = 3×X×I²

Where X is the reactance of the line.

Qloss = 3×j2.7×100²

Qloss = j81000 var

Qloss = j81 kvar

Sloss = Ploss + Qloss

Sloss = 12 + j81 kVA

c. Real power and reactive power supplied at the sending end of the line

The complex power at sending end of the line is

Ss = 3×Vs×I*

Ss = 3×(2401.7 < 4.58)×(100 < 36.87°)

Ss = 540.046 + j476.95 kVA

So the sending end real power is

Ps = 540.046 kW

So the sending end reactive power is

Qs = j476.95 kvar

7 0

2 years ago