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MatroZZZ [7]
3 years ago
15

Technician A says that rear-wheel drive vehicles usually get better traction than front-wheel drive vehicles. Technician B says

that front-wheel drive vehicles get extremely good traction due to the weight of the engine and transaxle. Who is correct?
a

Technician A

b

Technician B

c

Both A & B

d

Neither A or B
Engineering
1 answer:
Alex Ar [27]3 years ago
6 0
Both a and b

Disclaimer! (90% sure)
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A Pelton wheel is supplied with water from a lake at an elevation H above the turbine. The penstock that supplies the water to t
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Answer:

Following are the proving to this question:

Explanation:

\frac{D_1}{D} = \frac{1}{(2f(\frac{l}{D}))^{\frac{1}{4}}}

using the energy equation for entry and exit value :

\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0  = \frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g}

where

\to p_0=p_1=0\\\\\to Z_0=Z_1=H\\\\\to v_0=0\\\\AV =A_1V_1 \\\\\to V=(\frac{D_1}{D})^2 V_1\\\\\to V^2=(\frac{D_1}{D})^4 V^{2}_{1}

         = (\frac{1}{(2f (\frac{l}{D} ))^{\frac{1}{4}}})^4\  V^{2}_{1}\\\\

         = \frac{1}{(2f (\frac{l}{D})  )} \  V^{2}_{1}\\

\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0  =\frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g} \\\\

\to 0+0+Z_0 = 0  +\frac{V^{2}_{1} }{2g} +Z_1+ f \frac{l}{D} \frac{\frac{1}{(2f(\frac{l}{D}))}\ V^{2}_{1}}{2g}   \\\\\to Z_0 -Z_1 = +\frac{V^{2}_{1}}{2g} \ (1+f\frac{l}{D}\frac{1}{(2f(\frac{l}{D}) )} )  \\\\\to H= \frac{V^{2}_{1}}{2g} (\frac{3}{2}) \\\\\to  \frac{V^{2}_{1}}{2g} = H(\frac{3}{2})

L.H.S = R.H.S

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[Find the attachment]

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3 years ago
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Answer:

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Explanation:

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Compute L, T, M, LC, and R and stations of the BC and EC for the circular curve with the given data of: I (delta) = 22°15′00" an
Mars2501 [29]

Answer:

L = 475.718

T = 240.89 ft

M = 23.0195

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Explanation:

See the attached file for the calculation.

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2. What is the Function of the Camshaft in an Internal Combustion Engine?
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Answer:

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Explanation:

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