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Alex777 [14]
3 years ago
11

A simple undamped spring-mass system is set into motion from rest by giving it an initial velocity of 100 mm/s. It oscillates wi

th a maximum amplitude of 10 mm. What is its natural frequency?
Engineering
1 answer:
____ [38]3 years ago
6 0

Answer:

f=1.59 Hz

Explanation:

Given that

Simple undamped system means ,system does not consists any damper.If system consists damper then it is damped spring mass system.

Velocity = 100 mm/s

Maximum amplitude = 10 mm

We know that for a simple undamped system spring mass system

V_{max}=\omega A

now by putting the values

V_{max}=\omega A

100 = ω x 10

ω = 10 rad/s

We also know that

ω=2π f

10 = 2 x π x f

f=1.59 Hz

So the natural frequency will be f=1.59 Hz.

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An electric power plant uses solid waste for fuel in the production of electricity. The cost Y in dollars per hour to produce el
andrew-mc [135]

Answer:

15.64 MW

Explanation:

The computation of value of X that gives maximum profit is shown below:-

Profit = Revenue - Cost

= 15x - 0.2x 2 - 12 - 0.3x - 0.27x 2

= 14.7x - .47x^2 - 12

After solving the above equation we will get maximum differentiate  for profit that is

14.7 - 0.94x = 0

So,

x = 15.64 MW

Therefore for computing the value of X that gives maximum profit we simply solve the above equation.

8 0
3 years ago
Give me source code of Simple openGL project. ( without 3D or Animation) simple just.
Ivan

Answer:

Use GitHub or stackoverflow for this answer

Explanation:

It helps with programming a lot

4 0
3 years ago
For which of 'water' flow velocities at 200C can we assume that the flow is incompressible ? a.1000 km per hour b. 500 km per ho
ad-work [718]

Answer:d

Explanation:

Given

Temperature=200^{\circ}\approc 473 K

Also \gamma for air=1.4

R=287 J/kg

Flow will be In-compressible when Mach no.<0.32

Mach no.=\frac{V}{\sqrt{\gamma RT}}

(a)1000 km/h\approx 277.78 m/s

Mach no.=\frac{277.78}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.63

(b)500 km/h\approx 138.89 m/s

Mach no.=\frac{138.89}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.31

(c)2000 km/h\approx 555.55 m/s

Mach no.=\frac{555.55}{\sqrt{1.4\times 287\times 473}}

Mach no.=1.27

(d)200 km/h\approx 55.55 m/s

Mach no.=\frac{55.55}{\sqrt{1.4\times 287\times 473}}

Mach no.=0.127

From above results it is clear that for Flow at velocity 200 km/h ,it will be incompressible.

5 0
3 years ago
You must signal [blank] before any turn or lane change.
Soloha48 [4]
A. 5 seconds :) Good luck!
5 0
3 years ago
A fan that consumes 20 W of electric power when operating is claimed to discharge air from a ventilated room at a rate of 1.0 kg
solong [7]

Answer:

No, the claim is not reasonable for 20 W electric power consumption.

It is reasonable for 40 W electric power consumption.

Explanation:

Power = (1/2)*mass flow rate*(square of velocity)

mass flow rate = 1 kg/s

velocity = 8 m/s

square of velocity = 64 m^2 / s^2

Power = (1/2)*(1)*(64)

Power  = 32 W

For a fan that consumes 20 W power it is not possible to deliver more power than 20 W but this one is delivering 32 W hence it is a false claim.

For a fan that consumes 40 W it is indeed possible to deliver 32 W considering the efficiency. Hence this claim is reasonable.

5 0
3 years ago
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