Ask question

Ask question

All categories

jek_recluse [69]

1 year ago

13

if you run off the pavement, you should: turn the steering wheel quickly toward the road steer straight and slow down before att

empting to return to the pavement steer straight ahead and speed up apply the brakes hard submit answer

1 answer:

omeli [17]1 year ago

5 0

pavement is defined as the surface of Road or sidewalk.

for example, the surface of Expressway.

There are two types of pavement.

rigid pavement which consists of one layer.

flexible pavement which consist of multiple layers.

While driving on roads of rural areas, if our right wheel moves off the pavement, we should always hold the steering wheel firmly and then take our foot off the pedal, then apply brake lightly until we are moving at a low speed.

if you run off the pavement, you should: turn the steering wheel quickly toward the road steer straight and slow down before attempting to return to the pavement steer straight ahead and speed up apply the brakes hard

To know more about pavement:

brainly.com/question/28456065

#SPJ4

You might be interested in

A 4 kg object moving to the left collides with and sticks to a 3 kg object moving to the right. Which of the following is true o

madam [21]

Answer:

D. The motion cannot be determined without knowing the speeds of the objects before the collision.

Explanation:

This question is tricky! We know the object moving to the left has a greater mass than the one moving to the right. We'd <em>assume</em> they would move to the left because the leftwards object has a greater mass, right?

Not. So. Fast.

We can solve for the objects' final velocity using the formula for momentum, m₁v₁ + m₂v₂ = (m₁ + m₂)v .

Now here's where the trap is sprung: <em>we don't think about the equation</em>. This shows that the final velocity of the objects and the direction depends on both the mass of the objects <em>and</em> their initial velocity.

Basically, what if the 3 kg object is moving at 1 m/s and the 4 kg object is moving at –0.5 m/s? The objects would move to the <em>right</em> after the collision!

Do we know the velocity of these objects? No, right?

That means we <em>can't</em> determine the direction of their motion <u>unless we know their initial, pre-collision velocity</u>. This question is tricky because we look at the 4 kg vs. 3 kg and automatically assume the 4 kg object would dictate the direction of motion. That's not true. It depends on velocity as well.

I hope this helps you! Have a great day!

4 0

3 years ago

you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k

alex41 [277]

Answer:

You should choose airspeed 158.11 km/h at 18.4° west of south

Explanation:

The distance to the air port is 450 km due to south

You should to reach the airport in 3 hours

→ Velocity = distance ÷ time

→ Distance = 450 km , time = 3 hours

→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south

A wind is blowing from west 50 km/h

We need to know what heading and airspeed you should choose to

reach your destination

At first we must find the resultant velocity of your plane and the wind

The south and west are perpendicular, then the resultant velocity is

→ $v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}$

→ $v_{p}=150$ km/h , $v_{w}=50$ km/h

→ $v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11$ km/h

To cancel the velocity of the wind, the pilot should maintain the velocity

of the plane at 158.11 km/h

The direction of the velocity is the angle between the resultant velocity

and the vertical (south)

→ The direction of the velocity is $tan^{-1}\frac{50}{150}=18.4$ °

The direction of the velocity is 18.4° west of south

<em>You should choose airspeed 158.11 km/h at 18.4° west of south</em>

8 0

3 years ago

Explain what the star looks like to Jane

S_A_V [24]

The star looks like a desirable hunk of masculinity to Jane. But to John, the star looks like a wimpy momma's boy who might compete with him for Jane's attention. Jane and John have different impressions of the star because of their gender-specific instincts that have evolved during thousands of millenia of human evolution.

7 0

3 years ago

Read 2 more answers

The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.

Elena L [17]

Answer:

The magnitude of the acceleration is $a_r = 1.50 \ m/s^2$

The direction is $\theta = 32.5 6^o$ north of east

Explanation:

From the question we are told that

The force exerted by the wind is $F_{sail} = (330 ) \ N \ north$

The force exerted by water is $F_{keel} = (210 ) \ N \ east$

The mass of the boat(+ crew) is $m_b = 260 \ kg$

Now Force is mathematically represented as

$F = ma$

Now the acceleration towards the north is mathematically represented as

$a_n = \frac{F_{sail}}{m_b}$

substituting values

$a_n = \frac{330 }{260}$

$a_n = 1.269 \ m/s^2$

Now the acceleration towards the east is mathematically represented as

$a_e = \frac{F_{keel}}{m_b }$

substituting values

$a_e = \frac{210}{260}$

$a_e =0.808 \ m/s^2$

The resultant acceleration is

$a_r = \sqrt{a_e^2 + a_n^2}$

substituting values

$a_r = \sqrt{(0.808)^2 + (1.269)^2}$

$a_r = 1.50 \ m/s^2$

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

$tan \theta = \frac{a_e}{a_n}$

$\theta = tan ^{-1} [\frac{a_e}{a_n } ]$

substituting values

$\theta = tan ^{-1} [\frac{0.808}{1.269 } ]$

$\theta = tan ^{-1} [0.636 ]$

$\theta = 32.5 6^o$

5 0

3 years ago

Incident rays parallel to the axis of a concave mirror reflect parallel to the axis.

coldgirl [10]

No they don't. Incident rays parallel to the axis of a concave mirror
reflect from the mirror's surface and converge at its focal point.

3 0

2 years ago