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Temka [501]
4 years ago
9

Two cars A and B leave an intersection at the same time. Car A travels west at an average speed of x miles per hour and car B tr

avels south at an average speed of y miles per hour. Write a program that prompts the user to enter the average speed of both the cars and the elapsed time (in hours and minutes) and outputs the (shortest) distance between the cars.
Engineering
1 answer:
svet-max [94.6K]4 years ago
7 0

Answer:

Distance between A & B, dAB, is hypotenus.

dAB²=dA²+dB²

A'=70 mph, B'=55mph, (convert t to hours)

dA=A'•t, dB=B'•t

dAB=√((70t)²+(55t)²)

Explanation:

Distance between A & B, dAB, is hypotenus.

dAB²=dA²+dB²

A'=70 mph, B'=55mph, (convert t to hours)

dA=A'•t, dB=B'•t

dAB=√((70t)²+(55t)²)

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A cylinder 8 in. in diameter and 3 ft long is concentric with a pipe of 8.25 in internal diameter. Between the cylinder and the
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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}

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Answer:

Option c)

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