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2 years ago

The force of impact is

1 answer:

3 0

In mechanics, an impact is a high force or shock applied over a short time period when two or more bodies collide. Such a force or acceleration usually has a greater effect than a lower force applied over a proportionally longer period.

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A freeway is being designed for a location in rolling terrain. The expected free-flow speed is 55 mi/h. During the peak hour, it

balu736 [363]

Answer:

Explanation:

We start calcualting the factor for heavy-vehicle, that is,

$f_{HV} = \frac{1}{1+P_T(E_T-1)+P_R(E_R-1)}$

$E_T = 2.5$ for rolling terrain, so, replacing,

$f_{HV} = \frac{1}{1+0.18(2.5-1)+0}$

$f_{HV} = 0.7874$

We calculate the number of lanes by using the relation as follow:

$N=\frac{V}{\upsilon_p PHF f_{HV} f_p}$

$N=\frac{2700}{1915*0.88*0.7874*1}$

$N= 2.035$

In this way the necessary number of lanes is 3.

Note: To calculate $\upsilon_p$ you need to consult the Exhibit 23-2

in 'Highway capacity Manual' for 55mph and the LOS D criteria.

6 0

4 years ago

A condenser accepts steam from the turbine in problem 2 at a pressure of 2.34 kPa. Saturated water at the same pressure leaves t

vaieri [72.5K]

Answer:

The answer is "83.98, 1889.195, and 1889.195"

Explanation:

Given value:

$\bold{P_{4}=2.34 \ kPa}$

In point a:

The value of $h_{f4}=83.915 \ \ \frac{Kj}{kg}\\$

$V_4=0.001002 \ \ \frac{Kj}{kg}\\\\U_4= 83.98 \ \ \frac{Kj}{kg}\\\\$

In point b:

calculating heat leaves formula= $h_3-h_{f4}$

$= 1973.11-83.915\\\\= 1889.195 \ \ \frac{KJ}{kg}$

In point c:

calculating Heat transfer rate formula $=m(h_3-h_4)$

$= 1(1889.195)\\\\ = 1889.19 \ \ kw.$

7 0

3 years ago

A manager who focuses on the employees who enable a company to do business is human resource management. True True False False

nydimaria [60]

Answer:

False.

Explanation:

6 0

3 years ago

A customer complains that the heater works sometimes, but sometimes only cold air comes out while driving. Technician A says tha

murzikaleks [220]

Answer:

Technition A is correct

Explanation:

7 0

3 years ago

A polyethylene rod exactly 10 inches long with a cross-sectional area of 0.04 in2 is used to suspend a weight of 358 lbs-f (poun

Nadya [2.5K]

Answer:

Final length of the rod = 13.90 in

Explanation:

Cross Sectional Area of the polythene rod, A = 0.04 in²

Original length of the polythene rod, l = 10 inches

Tensile modulus for the polymer, E = 25,000 psi

Viscosity, $\eta = 1*10^{9} psi -sec$

Weight = 358 lbs - f

time, t = 1 hr = 3600 sec

Stress is given by:

$\sigma = \frac{Force}{Area} \\\sigma = \frac{358}{0.04} \\\sigma = 8950 psi$

Based on Maxwell's equation, the strain is given by:

$strain = \sigma ( \frac{1}{E} + \frac{t}{\eta} )\\Strain = 8950 ( \frac{1}{25000} + \frac{3600}{10^{9} } )\\Strain = 0.39022$

Strain = Extension/(original Length)

0.39022 = Extension/10

Extension = 0.39022 * 10

Extension = 3.9022 in

Extension = Final length - Original length

3.9022 = Final length - 10

Final length = 10 + 3.9022

Final length = 13.9022 in

Final length = 13.90 in

7 0

3 years ago