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murzikaleks [220]
3 years ago
11

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

light lasts only 2.0 s before turning to red, and she is 30 m away from the near side of the intersection (Fig. 2-29). Should she try to stop, or should she make a run for it? The intersection is 15 m wide. Her car's maximum deceleration is -7.0 m/s2, whereas it can accelerate from 48 km/h to 70 km/h in 6.6 s. Ignore the length of her car and her reaction time. If she hits the brakes, how far will she travel before stopping?
Physics
1 answer:
trasher [3.6K]3 years ago
4 0

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

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7 0
3 years ago
Describe the water cycle process starting from an afternoon thunderstorm. There are many different variations that could happen.
spin [16.1K]

Explanation:

The water cycle basically involves five steps:

  • evaporation and transpiration ⇄
  • condensation, ⇄
  • precipitation, ⇄
  • runoff, ⇄
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After this, the water <em>runoffs </em><em>to the surface of the ground, on plants, into rocks, rivers, and lakes.</em>

Next, the <em>Infiltration process</em> enables the water on the ground surface to enter the soil some of which becomes groundwater.

The cycle begins again as the<em> </em><em>evaporation and transpiration</em> <em>process </em>begins, where the groundwater as a result of heat from the sun is taken back into the atmosphere, while water in plants by means of transpiration goes back <em>into the atmosphere</em>.

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3 0
3 years ago
Problem 1 Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed
ANTONII [103]

Answer:

markers are 29.76 m far apart in the laboratory

Explanation:

Given the data in the question;

speed of particle = 0.624c

lifetime = 159 ns = 1.59 × 10⁻⁷ s

we know that; c is speed of light which is equal to 3 × 10⁸ m/s

we know that

distance = vt

or s = ut

so we substitute

distance = 0.624c × 1.59 × 10⁻⁷ s

distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s

distance = 1.872 × 10⁸ m/s × 1.59 × 10⁻⁷ s

distance =  29.76 m

Therefore, markers are 29.76 m far apart in the laboratory

3 0
3 years ago
Based on the law of conservation of energy, which statement is correct?
Nitella [24]

Answer:

You kinda left out the options you want us to choose from.

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4 0
3 years ago
A golf ball (m = 59.2 g) is struck a blow that makes an angle of 50.5 ◦ with the horizontal. The drive lands 130 m away on a fla
fomenos

Answer:

The force is calculated as 338.66 N

Explanation:

We know that force is given by

\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}=\frac{m(v_{f}-v_{o})}{\Delta t}

We know that range of a projectile is given by

R=\frac{v_{o}^{2}sin(2\theta )}{g}

it is given that R=130 m applying values in the above equation we get

R=\frac{v_{o}^{2}sin(2\theta )}{g}\\\\v_{0}=\sqrt{\frac{Rg}{sin(2\theta )}}\\\\\therefore v_{o}=\sqrt{\frac{130\times 9.81}{sin(2\times 50.5_{o}} )}\\\\v_{o}=36.04m/s

Thus the force is obtained as

\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}=\frac{59.2\times 10^{}-3(36.04-0)}{6.3\times 10^{-3}}

Thus force equals F=338.66N

4 0
3 years ago
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