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3 years ago

When you encounter a yellow light as you approach an intersection what is the safest approach to take?

1 answer:

5 0

This isn't a physics question really, but you should slow down unless you don't believe you can stop in time or don't believe it is safe to stop quickly (say you see someone driving very close behind you and you don't think they're paying attention). In general, if you think that by staying at your current speed your back tires will cross the ending lines of the intersection by the time the light turns red, it is safe to go through the yellow light. However, this is a thing you'll develop a feel for as you're driving, when in doubt, just slow down, just watch slamming on your breaks if there is someone behind you, sometimes people will see the yellow light when they're behind you and they'll speed up behind you so they'll "make it" before the light turns yellow. While this isn't illegal, since they're not technically running the red light, you should never speed up going up to a yellow light, if you need to speed up to make it before it turns red, you shouldn't make the light, just stop, this is especially bad if someone is directly in front of you and is likely to stop at the yellow light while the person behind them speeds up. This causes a lot of accidents.

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What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential di

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What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 1.50×10^4v ?

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1 year ago

A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k

Luba_88 [7]

Answer:

The frictional force $F_{fri} =$ 6.446 N

The acceleration of the block a = 6.04 $\frac{m}{s^{2} }$

Explanation:

Mass of the block = 3.9 kg

$\theta = 40$ °

$\mu$ = 0.22

(a). The frictional force is given by

$F_{fri} = \mu R_{N}$

$R_{N} = mg \cos \theta$

$R_{N} =$ 3.9 × 9.81 × $\cos 40$

$R_{N} =$ 29.3 N

Therefore the frictional force

$F_{fri} =$ 0.22 × 29.3

$F_{fri} =$ 6.446 N

(b). Block acceleration is given by

$F_{net} = F - F_{fri}$

F = 30 N

$F_{fri}$ = 6.446 N

$F_{net}$ = 30 - 6.446

$F_{net}$ = 23.554 N

The net force acting on the block is given by

$F_{net} = ma$

23.554 = 3.9 × a

a = 6.04 $\frac{m}{s^{2} }$

This is the acceleration of the block.

8 0

2 years ago

Answer question quick

Anettt [7]

Answer:

Yes , insulation has no role

5 0

3 years ago

The process of generating an electric current from the motion of a conductor in a magnetic field is

TiliK225 [7]

Magnetism is the answer

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2 years ago

A person driving her car at 48 km/h approaches an intersection just as the traffic light turns yellow. She knows that the yellow

trasher [3.6K]

Answer:

She must stop the car before interception, distance traveled 12.66 m

Explanation:

We will take all units to the SI system

Vo = 48Km / h (1000m / 1Km) (1h / 3600s) = 13.33 m / s

V2 = 70 Km / h = 19.44 m / s

We calculate the distance traveled before stopping

X = Vo t + ½ to t²

Time is what it takes traffic light to turn red is t = 2.0 s

X = 13.33 2 + 1.2 (-7) 2²

X = 12.66 m

It stops car before reaching the traffic light turning to red

Let's analyze what happens if you accelerate, let's calculate the acceleration of the vehicle

V2 = Vo + a t2

a = (V2-Vo) / t2

a = (19.44-13.33) /6.6

a = 0.926 m / s2

This is the acceleration to try to pass the interception, now let's calculate the distance it travels in the time the traffic light changes from yellow to red (t = 2.0 s)

X = Vo t + ½ to t²

X = 13.33 2 + ½ 0.926 2²

X = 28.58 m

Since the vehicle was 30 m away, the interception does not happen

4 0

3 years ago