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3 years ago

9

When you reach a yield sign yield to cross traffic and ___ before you enter the intersection

2 answers:

yawa3891 [41]3 years ago

8 0

Answer:

Wain for a safe gap

Explanation:

Musya8 [376]3 years ago

6 0

When approaching a yield sign, drivers must give the right-of-way to traffic already in the lanes that they intend to enter or cross. Drivers should be prepared to stop when approaching a yield sign but may continue without stopping if there is no conflicting traffic.

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The heat transfer coefficient decreases with increasing x for both the laminar and turbulent regions a. True b. False

REY [17]

Answer:

A) True

Explanation:

Yes this is true when length is creases the heat transfer coefficient decease with length.

The heat transfer(h) coefficient is varying with x by given expression

For Laminar flow

$h \alpha \dfrac{1}{x^{\frac{1}{2}}}$

For turbulent flow

$h \alpha \dfrac{1}{x^{\frac{1}{5}}}$

But when flow is in transitional state the heat heat transfer(h) coefficient is increases with x.But for laminar as well as turbulent flow h is decrease when x increases.

3 0

3 years ago

Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,

Lyrx [107]

Answer:

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$

Explanation:

$v_0$ = Initial velocity of ball A

$v_A=v_0\cos45^{\circ}$

$v_B$ = Initial velocity of ball B = 0

$(v_A)_n'$ = Final velocity of ball A

$v_B'$ = Final velocity of ball B

$e$ = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

$mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

Coefficient of restitution is given by

$e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

$(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

$v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

Adding the above two equations we get

$2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0$

$\boldsymbol{\therefore v_B'=0.6364v_0}$

$(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0$

From the conservation of momentum along the plane of contact we have

$(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}$

$v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}$

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$ .

5 0

3 years ago

Technician A says a solenoid is not used on modern automobiles. Technician B says fuel injectors and starter motor solenoids are

julia-pushkina [17]

Answer:

ON THE BOTTOM

Explanation:

red cord plus black cord

5 0

3 years ago

You hang a heavy ball with a mass of 42 kg from a silver rod 2.7 m long by 1.9 mm by 2.6 mm. You measure the stretch of the rod,

nadezda [96]

Answer:

Explanation:

cross sectional area A = 1.9 x 2.6 x 10⁻⁶ m²

= 4.94 x 10⁻⁶ m²

stress = 42 x 9.8 / 4.94 x 10⁻⁶

= 83.32 x 10⁶ N/m²

strain = .002902 / 2.7

= 1.075 x 10⁻³

Young's modulus = stress / strain

= 83.32 x 10⁶ / 1.075 x 10⁻³

= 77.5 x 10⁹ N/m²

5 0

3 years ago

Consider the following ways of handling deadlock: (1) banker’s algorithm, (2) detect

Andrew [12]

Answer:

b

Explanation:

7 0

3 years ago