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Natalija [7]
4 years ago
6

The vertical and horizontal poles at the traffic-light assembly are erected first. Determine the additional force and moment rea

ctions at the base O caused by the addition of the three 86-lb traffic signals B, C, and D. Report your answers as a force magnitude and a moment magnitude.
Engineering
1 answer:
konstantin123 [22]4 years ago
6 0

Answer:

1. Az=258 lb

2. My=3440 ft lb

3. ∑Mz= 0

Explanation:

∑ Fx= Ax=0

∑ Fy= Ay=0

∑ Fz= Az-86-86-86

Az=258 lb

∑Mx=86 X 31 +Mx=0

Mx=2666 ft lb

∑My=86 X 40 +My=0

My=3440 ft lb

∑Mz= 0

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Slav-nsk [51]

Answer:

The answer is A. that is, a merchant account allows you to use SSL on your website.

Explanation:

SSL means Secure Sockets Layer and this is an encryption-based Internet security protocol.

For an e-commerce merchant website or account, it is advised that an SSL package be installed to prevent any potential loss of private information.

For this reason, a merchant account allows use of SSL on your website because this also boost the confidence of client and customers visiting the website to purchase products.

7 0
3 years ago
A 1-ft rod with a diameter of 0.5 in. is subjected to a tensile force of 1,300 lb and has an elongation of 0.009 in. The modulus
iragen [17]

Answer:

E = 8.83 kips

Explanation:

First, we determine the stress on the rod:

\sigma = \frac{F}{A}\\\\

where,

σ = stress = ?

F = Force Applied = 1300 lb

A = Cross-sectional Area of rod = 0.5\pi \frac{d^2}{4} = \pi \frac{(0.5\ in)^2}{4} = 0.1963\ in^2

Therefore,

\sigma = \frac{1300\ lb}{0.1963\ in^2} \\\\\sigma = 6.62\ kips

Now, we determine the strain:

strain = \epsilon = \frac{elongation}{original\ length} \\\\\epsilon = \frac{0.009\ in}{12\ in}\\\\\epsilon =  7.5\ x\ 10^{-4}

Now, the modulus of elasticity (E) is given as:

E = \frac{\sigma}{\epsilon}\\\\E = \frac{6.62\ kips}{7.5\ x\ 10^{-4}}

<u>E = 8.83 kips</u>

7 0
3 years ago
Two transmission belts pass over a double-sheaved pulley that is attached to an axle supported by bearings at A and D. The radiu
Fynjy0 [20]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

6 0
4 years ago
A very large plate is placed equidistant between two vertical walls. The 10-mm spacing between the plate and each wall is filled
Vikentia [17]

Answer:

Force per unit plate area is 0.1344 N/m^{2}

Solution:

As per the question:

The spacing between each wall and the plate, d = 10 mm = 0.01 m

Absolute viscosity of the liquid, \mu =1.92\times 10^{- 3} Pa-s

Speed, v = 35 mm/s = 0.035 m/s

Now,

Suppose the drag force that exist between each wall and plate is F and F' respectively:

Net Drag Force = F' + F''

F = \tau A

where

\tau = shear stress

A = Cross - sectional Area

Therefore,

Net Drag Force, F = (\tau ' +\tau '')A

\frac{F}{A} = \tau ' +\tau ''

Also

F = \frac{\mu v}{d}

where

\mu = dynamic coefficient of viscosity

Pressure, P = \frac{F}{A}

Therefore,

\frac{F}{A} = \frac{\mu v}{d} + \frac{\mu v}{d} = 2\frac{\mu v}{d}

\frac{F}{A} = 2\frac{1.92\times 10^{- 3}\times 0.035}{0.010} = 0.01344 N/m^{2}

8 0
3 years ago
Deep-ocean fish cannot survive in shallow water, because the water
shutvik [7]

Answer: 3. Test the prototype against the design criteria.

Explanation: Dovnovan should test his prototype before anything eles to work out any kinks in the design and write down the problems, and build another one before telling others about the project

6 0
4 years ago
Read 2 more answers
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