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3 years ago

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A motor is mounted on a platform that is observed to vibrate excessively at an operating speed of 6000 rpm producing a 250-N for

ce. Design a vibration absorber (undamped) to add to the platform. Note that in this case the absorber mass will only be allowed to move 2 mm because of geometric and size constraints.

1 answer:

vichka [17]3 years ago

4 0

Answer:

The amplitude of the absorbed mass can be found

for ka:

$X_{a} =0.002m=\frac{F_{0} }{K_{a} } =\frac{250}{K_{a} } =125000N/m$

now

$w^2=\frac{K_{a} }{m_{a} } \\m_{a} =\frac{K_{a} }{w^2} =\frac{125000}{[6000*2\pi /60]^2} =0.317kg$

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3 years ago

Determine the following parameters for the water having quality x=0.7 at 200 kPa:

ra1l [238]

Solution :

Given :

Water have quality x = 0.7 (dryness fraction) at around pressure of 200 kPa

The phase diagram is provided below.

a). The phase is a standard mixture.

b). At pressure, p = 200 kPa, T = $T_{saturated}$

Temperature = 120.21°C

c). Specific volume

$v_{f}= 0.001061, \ \ v_g=0.88578 \ m^3/kg$

$v_x=v_f+x(v_g-v_f)$

$=0.001061+0.7(0.88578-0.001061)$

$=0.62036 \ m^3/kg$

d). Specific energy ( $u_x$ )

$u_f=504.5 \ kJ/kg, \ \ u_{fg}=2024.6 \ kJ/kg$

$u_x=504.5 + 0.7(2024.6)$

$=1921.72 \ kJ/kg$

e). Specific enthalpy $(h_x)$

At $h_f = 504.71, \ \ h_{fg} = 2201.6$

$h_x=504.71+(0.7\times 2201.6)$

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3 years ago

It is said that Archimedes discovered his principle during a bath while thinking about how he could determine if King Hiero’s cr

arlik [135]

Answer: The crown is not made of pure gold.

Explanation:

Archimedes discovered that any solid, of any shape, when submerged in a liquid receives an upward force, equal to the weight of the volume of the liquid removed by the solid, which is equal to the solid volume.

So, if any body is weighed in air, the normal force will be equal to the gravity force (which we call weight) which can be expressed as follows:

Fg = m g = δ V g = 34.7 N

When submerged in water, the normal force is equal to the difference between the actual weight, and the upward force due to Archimedes' principle, called buoyant force, as follows:

Fn = Fg - Ep = δx. V. g - δH20 . V. g = 31.5 N

Dividing Fg between Fn, and simplifying common terms, we have:

δx / (δx - δh20) = 34.7 / 31.5 = 1.10

Solving for δx, we get the following value:

δx = 11,000 Kg/m3, less dense than pure gold, so we can conclude that the crown was not made of pure gold.

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3 years ago

Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you

viktelen [127]

Answer:

$\theta_1=15^o\\\theta_2=75^o$

Explanation:

<u>Projectile Motion</u>

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

$V_x=V_{ox}=V_ocos\theta$

Where $\theta$ is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

$V_y=V_{oy}-gt=V_osin\theta-gt$

The horizontal and vertical distances are, respectively:

$x=V_{o}cos\theta t$

$\displaystyle y=y_o+V_{o}sin\theta t-\frac{gt^2}{2}$

The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

$\displaystyle t_f=\frac{2V_osin\theta}{g}$

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

$\displaystyle R=\frac{V_o^2sin2\theta}{g}$

Let's solve for $\theta$

$\displaystyle sin2\theta=\frac{R.g}{V_o^2}$

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:

$\displaystyle \theta_1=\frac{asin\left(\frac{R.g}{V_o^2}\right)}{2}$

$\displaystyle \theta_2=\frac{180^o-asin\left(\frac{R.g}{V_o^2}\right)}{2}$

Or equivalently:

$\theta_2=90^o-\theta_1$

Given Vo=37 m/s and R=70 m

$\displaystyle \theta_1=\frac{asin\left(\frac{70\times 9.8}{37^2}\right)}{2}$

$\theta_1=15^o$

And

$\theta_2=90^o-15^o=75^o$

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