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Pie
3 years ago
13

A 30-ft tall drop tower is being used to study the shape of water droplets as they fall through air. A camera is to be carried b

y a cam-operated linkage which will track the droplets motion from the 7-ft to the 9-ft point in its fall (measured from release point at the top of the tower). The drops are released every 0.5 sec. Every drop is to be filmed. Design a cam and follower which will track these droplets, matching their velocities and accelerations in the 1-ft filming window. Consider acceleration due to gravity, g = 32.2 ft/sec2 . Note that the follower is operating between 7-ft to the 9-ft but it follows the water droplets exactly within the 1-ft filming window.
Engineering
1 answer:
sveta [45]3 years ago
8 0

Answer:

The camera would flow the same velocity and acceleration of the droplet until 9ft. After that it would switch gear and accelerate back to its original spot at the rate of 140.54 m/s2 in order to catch the next droplet.

Explanation:

The time it takes for the drop to reach 7-ft and 9-ft at the acceleration of 32.2ft/s2:

s = gt^2/2

t^2 = 2s/g

t = \sqrt{2s/g}

When s = 7ft, t_7 = \sqrt{2*7/32.2} = 0.66 s

When s = 9ft, t_9 = \sqrt{2*9/32.2} = 0.75 s

So it would take Δt = 0.75 - 0.66 = 0.09 s for each drop to travel from 7ft to 9ft. As the drops are released every 0.5s, this means that after our cam follow the first drop to the end of the 7-9ft window, it has 0.5 - 0.09 = 0.41 s or less to reverse its acceleration and go back to the original spot to take care of the next drop.

So at 9ft, the drop and camera velocity would be:

v_9 =gt_9 = 32.2*0.75 = 24.1 ft/s

So if the camera reverses its direction to go back to original spot at the rate of a (ft/s2) and initial speed of 24.1 ft/s. Within the time span of 0.41s to catch the next drop at 7ft

7 = 9 + 24.1*(0.41) - a0.41^2/2

0 = 2 + 9.881 - 0.085a

a = 11.881 / 0.085 = 140.54 m/s^2

So the camera would flow the same velocity and acceleration of the droplet until 9ft. After that it would switch gear and accelerate back to its original spot at the rate of 140.54 m/s2 in order to catch the next droplet.

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Answer:

S = 172.69 J/K

Explanation:

For a closed system, the entropy is given as the natural logarithm of microstates.

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Entropy is the logarithm of number of microstates; Thus,

S = In ψ

However, the number of microstates is given by the formula;

ψ = Mⁿ

Thus, S = In Mⁿ

Plugging in the relevant values to obtain;

S = In (1000)^(25)

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A mass of 1.9 kg of air at 120 kPa and 24°C is contained in a gas-tight, frictionless piston–cylinder device. The air is now com
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Answer:

W=-260.66 kJ (negative answer means, that the work was done on the gas)

Explanation:

1) Convert temperature from C to K- T=24+273=297K- all temperature in the gas problems should be used in Kelvins;

2) We need to analyse type of the process- it is given, that the temperature is constant, so it is an Isothermal process, which means, that the equation of the process is: pV=const (constant);

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4) To calculate initical and final volumes (Va and Vb), we can use the following equation: pV=mRT, so V=mRT/p. Note, that the pressure is changing, thus we can calculate volumes for the both cases- initial and final, using initial (120kPa) and final (600kPa) pressures, in addition, we can find equation for the pressure, as function of the volume, which we need to use for the integration in step 3: p=mRT/V;

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Note, that natural logarithm (ln) yields negative answer, which supports the question, that the work was done on the gas, not by the gas.

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Answer:

1121.7 × 10³⁰ photons per second

Explanation:

Data provided in the question:

Power transmitted by the AM radio,P = 550 kW = 550 × 10³ W

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Now,

P = \frac{NE}{t}

here,

N is the number of photons

t is the time

E = energy = hf

h = plank's constant = 6.626 × 10⁻³⁴ m² kg / s

Thus,

P = \frac{NE}{t} = \frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}          [t = 1 s for per second]

or

550 × 10³ = \frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}

or

550 = N × 4903.24 × 10⁻³⁴

or

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3 years ago
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