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Digiron [165]
3 years ago
14

The arm of the robot is extending at a constant rate = 1.5 ft/s when r = 3 ft, z = (4t2) ft, and  = (1.5 t) rad, where t is in

seconds. Find: The velocity and acceleration of the grip A when t = 3 s.

Engineering
1 answer:
vlabodo [156]3 years ago
4 0

Answer:

Velocity of the grip A= 24.077 ft/sec ,at t=3 seconds

Acceleration of the grip A=8.2764 ft/sec^2, at t=3 seconds

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A civil engineer is asked to design a curved section of roadway that meets the following conditions: With ice on the road, when
lianna [129]

Answer:

1. 3.4^{o}

2. 163.3 m

Explanation:

Static friction between road and rubber, μs =0.06

The maximum speed of the car, v = 50 km/h

                                              = (50)(1000/3600) m/s

                                               = 13.89 m/s

The acceleration due to gravity, g = 9.81 m/s^{2}

The frictional force, f = μsN     ...... (1)

The component mg cosθ which balance the normal reaction N

The component mg sinθ acts in an opposite direction to the frictional force f.

        ΣF = mg sinθ-f = 0      ...... (2)

Substitute the equation (1) in equation (2), we get

 ΣF = mgsinθ-μsN = 0

 mgsinθ-μsmgcosθ =0

 μs = sinθ/cosθ

   tanθ = μs

    θ = tan-1( μs) = tan-1(0.06) = 3.4^{o}

(b)The vertical component of the force is

N cosθ = fsinθ+mg

 N cosθ = μsNsinθ+mg

N[cosθ- μs sinθ] = mg     ...... (3)

The horizontal component of the force along the motion of the car is

Nsinθ+fcosθ = ma  (Centripetal acceleration, a = \frac {v^{2}}{r}

  Nsinθ+fcosθ = m(\frac {v^{2}}{r})

   Nsinθ+μsNcosθ = m(\frac {v^{2}}{r})

N[sinθ+μs cosθ] = m(\frac {v^{2}}{r})     ...... (4)    

Dividing the equation (4) with equation (3),

 [sinθ+μscosθ]/[cosθ- μs sinθ] = \frac {v^{2}}{rg}

 cosθ[sinθ/cosθ+μs]/cosθ[1- μs sinθ/cosθ] =\frac {v^{2}}{rg}

[tanθ+μs]/[1-μs tanθ] = \frac {v^{2}}{rg}      

 From part (1), tanθ = μs

 Then the above equation becomes

 \frac {(\mu_s+\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}

\frac {(2\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}

Therefore, the minimum radius of the curvature of the curve is

               r = \frac {v^{2}}{{2 \mu_s/[1-\mu_s^{2}]}g} 

                   = \frac {v^{2}[1-\mu_s^{2}]}{2\mu_s g}

                   = \frac {(13.89 m/s)^{2}[1-(0.06)^{2}]}{(2)(0.06)(9.81)}

                 = 163.3 m

5 0
3 years ago
Methane and oxygen react in the presence of a catalyst to form formaldehyde. In a parallel reaction, methane is oxidized to carb
Nezavi [6.7K]

Answer:

y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02

Explanation:

Hello,

a. On the attached document, you can see a brief scheme of the process. Thus, to know the degrees of freedom, we state the following unknowns:

- \xi_1 and \xi_2: extent of the reactions (2).

- F_{O_2}^2, F_{CH_4}^2, F_{H_2O}^2, F_{HCHO}^2 and F_{CO_2}^2: Molar flows at the second stream (5).

On the other hand, we've got the following equations:

- F_{O_2}^2=50mol/s-\xi_1-2\xi_2: oxygen mole balance.

- F_{CH_4}^2=50mol/s-\xi_1-\xi_2: methane mole balance.

- F_{H_2O}^2=\xi_1+2\xi_2: water mole balance.

- F_{HCHO}^2=\xi_1: formaldehyde mole balance.

- F_{CO_2}^2=\xi_2: carbon dioxide mole balance.

Thus, the degrees of freedom are:

DF=7unknowns-5equations=2

It means that we need two additional equations or data to solve the problem.

b. Here, the two missing data are given. For the fractional conversion of methane, we define:

0.900=\frac{\xi_1+\xi_2}{50mol/s}

And for the fractional yield of formaldehyde we can set it in terms of methane as the reagents are equimolar:

0.860=\frac{F_{HCHO}^2}{50mol/s}

In such a way, one realizes that the output formaldehyde's molar flow is:

F_{HCHO}^2=0.860*50mol/s=43mol/s

Which is equal to the first reaction extent \xi_1, therefore, one computes the second one from the fractional conversion of methane as:

\xi_2=0.900*50mol/s-\xi_1\\\xi_2=0.900*50mol/s-43mol/s\\\xi_2=2mol/s

Now, one computes the rest of the output flows via:

- F_{O_2}^2=50mol/s-43mol/s-2*2mol/s=3mol/s

- F_{CH_4}^2=50mol/s-43mol/s-2mol/s=5mol/s

- F_{H_2O}^2=43mol/s+2*2mol/s=47mol/s

- F_{HCHO}^2=43mol/s

- F_{CO_2}^2=2mol/s

The total output molar flow is:

F_{O_2}+F_{CH_4}+F_{H_2O}+F_{HCHO}+F_{CO_2}=100mol/s

Therefore the output stream composition turns out into:

y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02

Best regards.

7 0
2 years ago
19. A circuit contains four 100 S2 resistors connected in series. If you test the circuit with a digital VOM,
yanalaym [24]

Answer:

D

Explanation:

in series circuit the resistance is divided Total resistance is equal to the sum of resistances

6 0
2 years ago
Twenty-five wooden beams were ordered or a construction project. The sample mean and he sample standard deviation were measured
aksik [14]

Answer:

Correct option: B. 90%

Explanation:

The confidence interval is given by:

CI = [\bar{x} - z\sigma_{\bar{x}} , \bar{x}+z\sigma_{\bar{x}} ]

If \bar{x} is 190, we can find the value of z\sigma_{\bar{x}}:

\bar{x} - z\sigma_{\bar{x}}  = 188.29

190 - z\sigma_{\bar{x}}  = 188.29

z\sigma_{\bar{x}}  = 1.71

Now we need to find the value of \sigma_{\bar{x}}:

\sigma_{\bar{x}} = s / \sqrt{n}

\sigma_{\bar{x}} = 5/ \sqrt{25}

\sigma_{\bar{x}} = 1

So the value of z is 1.71.

Looking at the z-table, the z value that gives a z-score of 1.71 is 0.0436

This value will occur in both sides of the normal curve, so the confidence level is:

CI = 1 - 2*0.0436 = 0.9128 = 91.28\%

The nearest CI in the options is 90%, so the correct option is B.

4 0
3 years ago
Answer this question fast
laila [671]

Answer:

When the expenditure increased, then the consumer's expenditure is increased to Rs.150 and when the price falls of the good it becomes Rs.5. Then, Good X will be Rs.10.

last question:If the Good X falls by 20%, then, it will be Rs.2, and according to his demand 100 units will be equal to Rs.200.

cause if one unit=rs.2, then 100units=100×2=200.

8 0
3 years ago
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