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Elena L [17]
3 years ago
12

Our rule-of-thumb for presenting final results is to round to three significant digits or four if the first digit is a one. By t

his rule, write the result for
σ = Me/I , where M= 1835 lbf in and I/c = 0.875 in^3
Engineering
1 answer:
olga_2 [115]3 years ago
4 0

Answer:

To four significant digits = 2097 psi

Explanation:

<u>Applying the rule of thumb </u>

σ = Mc/I  ---- ( 1 )

M = 1835 Ibf in ,  I/c = 0.875 in^3

∴ c/l = 1 / 0.875 = 1.1429

back to equation 1

σ = 1835 * 1.1429 = 2097.2215 psi

To four significant digits = 2097 psi

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An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e
Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

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The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in 2.25 * 10^4 BTU. It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.  

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

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2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:  

Ep = m*g*(h_2 - h_1)\\ W = m*g  

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.  

h_2 is the final elevation and h_1 is the initial elevation.

Replacing W in the Ep equation

Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\

Finally, the next step is to replace the variables of the problem.  

h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft

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3 years ago
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Answer:

both statement is correct

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