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tangare [24]
2 years ago
9

Air is contained in a vertical piston–cylinder assembly such that the piston is in static equilibrium. The atmosphere exerts a p

ressure of 101 kPa on top of the 0.5-m-diameter piston. The gage pressure of the air inside the cylinder is 1.2 kPa. The local acceleration of gravity is g = 9.81 m/s2 . Subsequently, a weight is placed on top of the piston causing the piston to fall until reaching a new static equilibrium position. At this position, the gage pressure of the air inside the cylinder is 2.8 kPa. Determine (a) the mass of the piston, in kg, and (b) the mass of the added weight, in kg
Engineering
1 answer:
oee [108]2 years ago
5 0

Answer:

a) 24 kg

b) 32 kg

Explanation:

The gauge pressure is of the gas is equal to the weight of the piston divided by its area:

p = P / A

p = m * g / (π/4 * d^2)

Rearranging

p * (π/4 * d^2) = m * g

m = p * (π/4 * d^2) / g

m = 1200 * (π/4 * 0.5^2) / 9.81 = 24 kg

After the weight is added the gauge pressure is 2.8kPa

The mass of piston plus addded weight is

m2 = 2800 * (π/4 * 0.5^2) / 9.81 = 56 kg

56 - 24 = 32 kg

The mass of the added weight is 32 kg.

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A Class A fire extingisher is for use on general combustibles such as:​
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Two rods, with masses MA and MB having a coefficient of restitution, e, move along a common line on a surface, figure 2. a) Find
ahrayia [7]

Answer:

A.) Find the answer in the explanation

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C.) Impulse = 17.6 Ns

D.) 49%

Explanation:

Let Ua = initial velocity of the rod A

Ub = initial velocity of the rod B

Va = final velocity of the rod A

Vb = final velocity of the rod B

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Mb = mass of rod B

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Please find the attached files for the solution

6 0
3 years ago
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5 0
3 years ago
A square-thread power screw is used to raise or lower the basketball board in a gym, the weight of which is W = 100kg. See the f
KIM [24]

Answer:

power = 49.95 W

and it is self locking screw

Explanation:

given data

weight W = 100 kg = 1000 N

diameter d = 20mm

pitch p = 2mm

friction coefficient of steel f = 0.1

Gravity constant is g = 10 N/kg

solution

we know T is

T = w tan(α + φ ) \frac{dm}{2}     ...................1

here dm is = do - 0.5 P

dm = 20 - 1

dm = 19 mm

and

tan(α) = \frac{L}{\pi dm}      ...............2

here lead L = n × p

so tan(α) = \frac{2\times 2}{\pi 19}

α = 3.83°  

and

f = 0.1

so tanφ = 0.1

so that φ = 5.71°

and  now we will put all value in equation 1 we get

T = 1000 × tan(3.83 + 5.71 ) \frac{19\times 10^{-3}}{2}  

T = 1.59 Nm

so

power = \frac{2\pi N \ T }{60}     .................3

put here value

power = \frac{2\pi \times 300\times 1.59}{60}

power = 49.95 W

and

as φ > α

so it is self locking screw

 

8 0
3 years ago
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