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3 years ago

Clutches And Torque Converters both provide all of these EXCEPT:

Engineering

1 answer:

DENIUS [597]3 years ago

5 0

Answer:

B I think

Explanation:

clutches has a fluid drive

both can stop and idle

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Using the AASHTO 1993 Flexible Pavement Design Procedure, design a pavement cross section that will provide 10 years service. Th

natali 33 [55]

Answer:

Check the explanation

Explanation:

Single Unit Truck ESAL = 43.38 + 5.16 = 48.54

Semi Unit Truck ESAL = 43.38+ 6.00+7.4 = 56.78

So total ESAL's during design life = (400*48.54 + 350*56.78)*365*10/18000 = (19416+19873)*3650= 3939

Kindly check the attached image

Here

Reliability = 95% = 0.95, therefore ZR = -1.645, S0 = 0.4, MR = 18

Delta PSI = 4.2-2.5= 1.7

Resilient Modulus = 18000 psi, So MR = 18

Assume SN = 3.0 for flexible pavements

There W18 calculates to 0.26807

log10 (3939) = 9.36*log10(SN+1) -.2/(.4+1094/(SN+1)5.19)) -6.01

Structural Number SN = a1*d1 + a2*d2 *m2 +a2*d3 *m3

= a1*d1 + a2*d2 +a2*d3

5 0

3 years ago

A body is moving with simple harmonic motion. It's velocity is recorded as being 3.5m/s when it is at 150mm from the mid-positio

natima [27]

Answer:

1) A=282.6 mm

2) $a_{max}=60.35\ m/s^2$

3)T=0.42 sec

4)f= 2.24 Hz

Explanation:

Given that

V=3.5 m/s at x=150 mm ------------1

V=2.5 m/s at x=225 mm ------------2

Where x measured from mid position.

We know that velocity in simple harmonic given as

$V=\omega \sqrt{A^2-x^2}$

Where A is the amplitude and ω is the natural frequency of simple harmonic motion.

From equation 1 and 2

$3.5=\omega \sqrt{A^2-0.15^2}$ ------3

$2.5=\omega \sqrt{A^2-0.225^2}$ --------4

Now by dividing equation 3 by 4

$\dfrac{3.5}{2.5}=\dfrac {\sqrt{A^2-0.15^2}}{\sqrt{A^2-0.225^2}}$

$1.96=\dfrac {{A^2-0.15^2}}{{A^2-0.225^2}}$

So A=0.2826 m

A=282.6 mm

Now by putting the values of A in the equation 3

$3.5=\omega \sqrt{A^2-0.15^2}$

$3.5=\omega \sqrt{0.2826^2-0.15^2}$

ω=14.609 rad/s

Frequency

ω= 2πf

14.609= 2 x π x f

f= 2.24 Hz

Maximum acceleration

$a_{max}=\omega ^2A$

$a_{max}=14.61 ^2\times 0.2826\ m/s^2$

$a_{max}=60.35\ m/s^2$

Time period T

$T=\dfrac{2\pi}{\omega}$

$T=\dfrac{2\pi}{14.609}$

T=0.42 sec

8 0

4 years ago

Create a series of eight successive displacements that would program a robot to move in an octagonal path that is as close as yo

Komok [63]

Answer:

bts biot bts biot jungkukkk

jungkukkkbiot

Explanation:

bts biot bts biot jungkukkk

jungkukkkbiot

5 0

3 years ago

While discussing run-flat tires: Technician A says that some are self-sealing tires and are designed to quickly and permanently

musickatia [10]

Answer:

The correct option is d ( Neither A nor B)

Explanation:

Technician A made 2 mistakes in his statement.Firstly the tire is self supporting not self sealing.

Secondly, this tire does not provide permanent sealing of punctured area option a is incorrect.

This self-supporting tire after being affected with complete air leakage can temporarily bear the load of the car and avoid rolling over a distance of 80 km at a maximum speed of 55 mph. Here is what technician B suggested incorrectly as the tire after being.Here the technician B suggested incorrectly as the tire after being affected with puncture can not travel at any speed so option B is wrong

Since option a and b are incorrect and c is invalid.

4 0

3 years ago

The shaft is made of A992 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allows free rotation.

zysi [14]

Answer:

the angle of twist of B with respect to D is -1.15°

the angle of twist of C with respect to D is 1.15°

Explanation:

The missing diagram that is supposed to be added to this image is attached in the file below.

From the given information:

The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.

For the Modulus of Rigidity G = 11 × 10³ Ksi = 11 × 10⁶ lb/in²

The objective are :

1) To determine the angle of twist of B with respect to D

Considering the Polar moment of Inertia at the shaft $J\tau$

shaft $J\tau$ = $\dfrac{\pi}{2}r^4$

where ;

r = 1 in /2

r = 0.5 in

shaft $J \tau$ = $\dfrac{\pi}{2} \times 0.5^4$

shaft $J\tau$ = 0.098218

Now; the angle of twist at B with respect to D is calculated by using the expression

$\phi_{B/D} = \sum \dfrac{TL}{JG}$

$\phi_{B/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

where;

$T_{CD} \ \ and \ \ L_{CD}$ are the torques at segments CD and length at segments CD

${T_{BC} \ \ and \ \ L_{BC}}$ are the torques at segments BC and length at segments BC

Also ; from the diagram; the following values where obtained:

$L_{BC}}$ = 2.5 in

$J\tau$ = 0.098218

G = 11 × 10⁶ lb/in²

$T_{BC$ = -60 lb.ft

$T_{CD$ = 0 lb.ft

$L_{CD$ = 5.5 in

$\phi_{B/D} = 0+ \dfrac{[(-60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{B/D} = \dfrac{[(-720 )] (30 )}{1079980}$

$\phi_{B/D} = \dfrac{-21600}{1079980}$

$\phi_{B/D} =$ − 0.02 rad

To degree; we have

$\phi_{B/D} = -0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{B/D} = -1.15^0}$

Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction

Thus; the angle of twist of B with respect to D is 1.15°

(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places

For the angle of twist of C with respect to D; we have:

$\phi_{C/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{C/D} = 0+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{B/D} = 0+ \dfrac{[(60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{C/D} = \dfrac{21600}{1079980}$

$\phi_{C/D} =$ 0.02 rad

To degree; we have

$\phi_{C/D} = 0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{C/D} = 1.15^0}$

3 0

3 years ago