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Y_Kistochka [10]
3 years ago
5

A developer has requested permission to build a large retail store at a location adjacent to the intersection of an undivided fo

ur-lane major road and a two-lane minor road. Traffic on the minor road is controlled by a stop sign. The speed limits are 35 and 50 mi/h on the minor and major roads, respectively. The building is to be located at a distance of 65 ft from the near lane of one of the approaches of the minor road. Determine the minimum sight distance required on the major road for the stopped vehicle on the minor road to safely turn right onto the major road. Design vehicle is a single-unit truck. Lanes on the major road are 12 ft wide. Hint: Adjust for the number of lanes. Given: Time gap for single-unit truck for Right turn = 8.5 sec Time adjustment for number of lanes = 0.7 sec
Engineering
1 answer:
Sergio [31]3 years ago
6 0

Answer:

676 ft

Explanation:

Minimum sight distance, d_min

d_min = 1.47 * v_max * t_total where v_max is maximum velocity in mi/h, t_total is total time

v_max is given as 50 mi/h

t_total is sum of time for right-turn and adjustment time=8.5+0.7=9.2 seconds

Substituting these figures we obtain d_min=1.47*50*9.2=676.2 ft

For practical purposes, this distance is taken as 676 ft

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(35-39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 68.0° F. Th
arlik [135]

Answer:

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

Explanation:

From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation (\dot Q), measured in BTU per hour, is represented by this formula:

\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4}) (1)

Where:

\epsilon - Emissivity, dimensionless.

A - Surface area of the student, measured in square feet.

\sigma - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

T_{s} - Temperature of the student, measured in Rankine.

T_{b} - Temperature of the bus, measured in Rankine.

If we know that \epsilon = 0.90, A = 16.188\,ft^{2}, \sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}, T_{s} = 554.07\,R and T_{b} = 527.67\,R, then the heat transfer rate due to electromagnetic radiation is:

\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}]

\dot Q = 417.492\,\frac{BTU}{h}

Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

Q = \dot Q \cdot \Delta t (2)

Where \Delta t is the heat transfer time, measured in hours.

If we know that \dot Q = 417.492\,\frac{BTU}{h} and \Delta t = \frac{1}{3}\,h, then the net energy transfer is:

Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)

Q = 139.164\,BTU

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

7 0
2 years ago
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Answer:

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Explanation:

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Answer:

Explanation:

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If a shear stress acts in one plane of an element, there must be an equal and opposite shear stress acting on a plane that is
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Answer:

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Explanation:

In the case when the sheer stress acts in the one plane of an element so it should be equal and opposite also the shear stress acted on a plan i.e. 90 degrees from the plane

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