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grigory [225]
2 years ago
4

Does the child in the crosswalk have the right of way? Explain why or why not.

Engineering
2 answers:
Solnce55 [7]2 years ago
4 0

Answer:

2.) The child in the crosswalk has the right of way because pedestrians always have right of way since they aren't in a vehicle. You must yield until the pedestrian has cleared the crosswalk even if the signal is green.

MARK ME BRAINLIST

yuradex [85]2 years ago
4 0

Answer:

When crossing a roadway pedestrians have the right-of-way when in a marked or an unmarked crosswalk. But they must yield to vehicles if the road isn't marked. They must only cross when the signal allows them.

Explanation:

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2. A ¼ in. diameter rod must be machined on a lathe to a smaller diameter for use as a specimen in a tension test. The rod mater
gladu [14]

Answer:

we use

sigma=force /area\\A=\pi r^2\\D=2*r

5 0
2 years ago
A thin, flat plate that is 0.2 m × 0.2 m on a side is oriented parallel to an atmospheric airstream having a velocity of 40 m/s.
Leto [7]

Answer:

The rate of heat transfer from both sides of the plate to the air is 240 W

Explanation:

Given;

area of the flat plate = 0.2 m × 0.2 m = 0.04 m²

velocity of atmospheric air stream, v = 40 m/s

drag force, F =  0.075 N

The rate of heat transfer from both sides of the plate to the air:

q = 2 [h'(A)(Ts -T∞)]

where;

h' is heat transfer coefficient obtained from Chilton-Colburn analogy

h' = \frac{C_f}{2} \rho u C_p P_r^{-2/3}\\\\\frac{C_f}{2} = \frac{\tau'_s}{2*\rho u^2/2}

Properties of air at 70°C and 1 atm:

ρ = 1.018 kg/m³, cp = 1009 J/kg.K, Pr = 0.7, v = 20.22 x 10⁻⁶ m²/s

\frac{C_f}{2} = \frac{(0.075/2)/(0.2)^2}{2*(1.018)(40)^2/2} = 5.756*10^{-4}\\\\Thus,\\h' = 5.756*10^{-4} (1.018*40*1009)*(0.7)^{-2/3}\\\\h' = 30 \ W/m^2.K

Finally;

q = 2 [ 30(0.04)(120 - 20) ]

q = 240 W

Therefore, the rate of heat transfer from both sides of the plate to the air is 240 W

4 0
2 years ago
Read 2 more answers
Q10. Select the correct option for the following questions – (10 points, 2 each) a. After an edge dislocation has passed through
Gnom [1K]

Answer:

a. True - Because the atomic arrangements of that region is disorderer because of the extra half plane atoms in between the line

b.Slip

C. Strength theoretical is greater than strength experimental

d. Shear stress

e. Highest linear density

4 0
2 years ago
1. A glass window of width W = 1 m and height H = 2 m is 5 mm thick and has a thermal conductivity of kg = 1.4 W/m*K. If the inn
emmasim [6.3K]

Answer:

1. \dot Q=19600\ W

2. \dot Q=120\ W

Explanation:

1.

Given:

  • height of the window pane, h=2\ m
  • width of the window pane, w=1\ m
  • thickness of the pane, t=5\ mm= 0.005\ m
  • thermal conductivity of the glass pane, k_g=1.4\ W.m^{-1}.K^{-1}
  • temperature of the inner surface, T_i=15^{\circ}C
  • temperature of the outer surface, T_o=-20^{\circ}C

<u>According to the Fourier's law the rate of heat transfer is given as:</u>

\dot Q=k_g.A.\frac{dT}{dx}

here:

A = area through which the heat transfer occurs = 2\times 1=2\ m^2

dT = temperature difference across the thickness of the surface = 35^{\circ}C

dx = t = thickness normal to the surface = 0.005\ m

\dot Q=1.4\times 2\times \frac{35}{0.005}

\dot Q=19600\ W

2.

  • air spacing between two glass panes, dx=0.01\ m
  • area of each glass pane, A=2\times 1=2\ m^2
  • thermal conductivity of air, k_a=0.024\ W.m^{-1}.K^{-1}
  • temperature difference between the surfaces, dT=25^{\circ}C

<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>

\dot Q=k_a.A.\frac{dT}{dx}

\dot Q=0.024\times 2\times \frac{25}{0.01}

\dot Q=120\ W

5 0
2 years ago
what is the advantage of decreasing the field current of a separately excited dc motor below its nominal value
enyata [817]

Answer:

Ability to rotate at higher speeds

Explanation:

Constant K1 becomes greater than the other constant K2

This translates to that the motor being able to rotate at high speeds, without necessarily exceeding the nominal armature voltage.

The armature voltage is the voltage that is developed around the terminals of the armature winding of an Alternating Current, i.e AC or a Direct Current, i.e DC machine during the period in which it tries to generate power.

6 0
2 years ago
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