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3 years ago

1. An arrow signal with a left-pointing arrow_

Engineering

2 answers:

Paul [167]3 years ago

6 0

Answer:b

Explanation:It’s only applies to the left-turning traffic

Whitepunk [10]3 years ago

5 0

Answer:

correct option is B. only applies to left-turning traffic

Explanation:

solution

according to traffic rule regulation , when we see signal with left pointing arrow it mean that we can move on the left side turn

and if we see left pointing signal with red arrow it meaning same as NO TURN ON RED

and when we see yellow arrow it also means the same as the yellow light

and if there is green arrow it mean we can freely now move on left

so here correct option is B. only applies to left-turning traffic

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Why does teachers grade things that are not due yet

pav-90 [236]

I think because if you’ve already turned it in they might as well grade asap instead of waiting

4 0

3 years ago

Read 2 more answers

Joe is a chemical engineer whose plant discharges heavy metals into the local river. By the test authorized by the city governme

chubhunter [2.5K]

Answer:

B probably

Explanation:

Because the prompt doesn't specify what sort of violation it could be anything maybe when they release the metals during the day and so on.

5 0

2 years ago

I'll mark brainliest plz help

Citrus2011 [14]

Answer:

Explanation:

There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

7 0

3 years ago

A steel pipe of 400-mm outer diameter is fabricated from 10-mm-thick plate by welding along a helix that forms an angle of 20° w

Verdich [7]

Explanation:

Outer di ameter $d_{0}=400 \mathrm{mm}[tex] Thickness of the cylinder [tex]t=10 \mathrm{mm}$

$\therefore[tex] Inner diam eter [tex]d_{i}=d_{0}-2 t=400-2 \times 10$

$d_{1}=380 \mathrm{mm}$

Given loading on the cylinder $P=300 \mathrm{kN}$ Helix an gle of the weld form $\theta=20^{\circ}$

(i) Normal stress on the plane at angle $\theta=20^{\circ}$ is

$\sigma=\frac{P \cos ^{2} \theta}{A_{0}}$

$\text { Where } A_{0}=\frac{\pi}{4}\left(d_{0}^{2}-d_{1}^{2}\right)$

$\quad=\frac{\pi}{4}\left(400^{2}-380^{2}\right)$

$=12252.21 \mathrm{mm}^{2}$

$=12.25221 \times 10^{-9} \mathrm{m}^{2}$

$\sigma=\frac{-300 \times 10^{2} \times \cos ^{2} 20}{12.25221 \times 10^{-1}}$

$=-21.6 \mathrm{MPa}$

(ii) Shear stress along an angle of $\theta=20^{\circ}$ is $\tau=\frac{P}{A_{0}} \cos \theta \sin \theta$

$=\frac{-300 \times 10^{-1} \times \cos 20 \times \sin 20}{12.25221 \times 10^{-3}}$

$=-7.86 \mathrm{MPa}$

3 0

2 years ago

The cable connecting the winch A to point B on the railroad car in figure is wound in at the constant rate of 2 m. When θ = 60◦

Crank

Answer:

This question comprises two independent parts, (a) and (b) (a) In the diagram below, the cable connecting the winch A with point B on the railway carriage is being wound in at a constant rate of 2 m/s. Determine, for the instant in time when the angle θ= 60° (i) The velocity vector of the railway carriage (ii) The acceleration vector of the railway carriage The radius of the winch can be considered negligible. (2 marks) (3 marks) (b) The diagram below shows two rotating ‘driving links, OA and BC, connected by a telescoping link AB. Determine the angular velocity vector of the telescoping link AB when it is in the position shown if the driving links have the angular velocities indicated in the diagram. 150 2 rad/s 45 165 Dimensions in millimeters 60 0 2 rad/s

7 0

2 years ago