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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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In an experiment to determine relative velocities of a feather, ball, and anvil, what is the object called?

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Independent variable if I’m not mistaken

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2 years ago

Assume you are an observer standing at a point along a three-lane roadway. All vehicles in lane 1 are traveling at 30 mi/h, all

garik1379 [7]

Answer:

Explanation:

Given data;

In line 1, v1 = 30mi/hr

in line 2, v2 = 45mi/hr

in line 3, v3 = 60mi/hr

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= VT = 45mi/hr

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2 years ago

The velocity of a particle which moves along the s-axis is given by v = 2-4t+5t^(3/2), where t is in seconds and v is in meters

Brut [27]

Answer:

a) s = 42 m

b) V = 26 m/s

c) a = 11 m/s²

Explanation:

The velocity(V) = 2 - 4t + 5t^3/2 where t is in seconds and V is in m/s

a) To get the position, we have to integrate the velocity with respect to time.

We get the position(s) from the equation:

$V=\frac{ds}{dt}$

$ds=Vdt$

Integrating both sides,

$s=\int\ {V} \, dt$

$s=\int\ {(2-4t+5t^{\frac{3}{2} }) \, dt$

Integrating, we get

$s=(2t-2t^{2} +2t^{\frac{5}{2} }+c)m$

But at t=0, s(0) = 2 m

Therefore,

$s(0)=2(0)-2(0)^{2} +2(0)^{\frac{5}{2} }+c$

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c = 2

Therefore

$s=(2t-2t^{2} +\frac{10}{5}t^{\frac{5}{2} }+2)m$

When t = 4,

$s=2(4)-2(4)^{2} +2(4)^{\frac{5}{2} }+2$

$s=8-32+64+2$

s = 42 m

At t= 4s, the position s = 42m

b) The velocity equation is given by:

$V=(2-4t+5t^{\frac{3}{2} })m/s$

At t = 4,

$V=2-4(4)+5(4)^{\frac{3}{2} }$

V = 2 - 16 + 40 = 26 m/s

The velocity(V)at t = 4 s is 26 m/s

c) The acceleration(a) is given by:

$a=\frac{dv}{dt}$

$a=\frac{d}{dt}(2-4t+5t^{ \frac{3}{2}})$

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$a=-4+\frac{15}{2}t^{\frac{1}{2} }$

$a=(-4+\frac{15}{2}t^{\frac{1}{2} })m/s^{2}$

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2 years ago

Read 2 more answers

# derive is a kind of which dependency

Gemiola [76]

Types of dependency relationships
Type of dependency Keyword or Stereotype
Abstraction «abstraction», «derive», «refine», or «trace»
Binding «bind»
Realization «realize»
Substitution «substitute

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2 years ago