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Ilya [14]
3 years ago
9

When Maggie applies the brakes of her car, the car slows uniformly from 15.3 m/s to 0 m/s in 2.23 s. How far ahead of a stop sig

n must she apply her brakes in order to stop at the sign? Answer in units of m
Physics
1 answer:
nirvana33 [79]3 years ago
3 0

Answer:

The answer is 34.119m

Explanation:

You need to pay attention to the word "uniformly". It means there is no acceleration thus the physics of this problem respond to the uniform rectilinear motion equations:

  • Xf = Xo + vt
  • v = constant
  • a = 0

where:

  • Xf = Final position
  • Xo = Origin position
  • v = velocity (speed)
  • t = time
  • a = acceleration

Xf = 0 + (15.3)(2.23)

Xf = 34.119m

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In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between st
bekas [8.4K]

Answer:

a) t = 746 s

b) t = 666 s

Explanation:

a)

  • Total time will be the sum of the partial times between stations plus the time stopped at the stations.
  • Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
  • At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
  • In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

       v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s  (1)

  • Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

       t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)

  • Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

       x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m  (3)

  • In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

       t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)

  • We can find the distance traveled while the train was decelerating as follows:

       x_{3} = (v_{1} * t_{3})   + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m  (5)

  • Finally, we need to know the time traveled at constant speed.
  • So, we need to find first the distance traveled at the constant speed of 26.4m/s.
  • This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
  • x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s   (7)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
  • Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
  • tm = t*5 = 131.6 * 5 = 658.2 s (9)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
  • Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

b)

  • Using all the same premises that for a) we know that the only  difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
  • Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
  • We can find the distance traveled at constant speed, rewriting (6) as follows:
  • x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s   (12)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
  • Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
  • tm = t*3 = 207.4 * 3 = 622.2 s (14)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
  • Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)
7 0
2 years ago
Ultrasonic images are obtained from the inside organs of our body. This process uses which property of sound wave?
Temka [501]

This question involves the concepts of echo, ultrasonic images, ultrasonic sound waves.

The process of ultrasonic images uses the "echo" property of the sound waves.

Echo is the property of the sound wave by the virtue of which the sound wave reflects back to the source of the sound after hitting a surface or an object.

Ultrasonic images are obtained from inside organs of our body. This process involves the use of ultrasonic sound waves that have a frequency greater than 20,000 Hz. These sound waves are out of the range of audible sound by the human ear. When these ultrasonic sound waves are sent in form of pulses into the human body by the use of probes, they reflect back from the tissues of different organs to the probe. The probe then records the reflection properties of these sound waves and displays them in form of an image, known as ultrasonic images.

Learn more about echo here:

brainly.com/question/14335186?referrer=searchResults

The attached picture shows the process of ultrasonic imaging.

4 0
2 years ago
During an observation, it was noticed that light diffracts as it passes through small slits in a barrier. What does this evidenc
Lynna [10]

Answer:

It reveals that light is a wave

Explanation:

Diffraction is the property of a wave in which there is a bending of the wave about the corners of an obstacle or aperture into the geometrical shadow of the obstacle or aperture.

This simply implies that a wave bends or spreads out when it passes through openings. Since the light diffracts through small slits and diffraction has been shown to occur in water waves and sound waves, this property of diffraction can only be characteristic of a wave and thus, this evidence reveals that light is a wave.

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A 1000 kg elevator accelerates upward at 1.0 m/s2 for 10 m, starting from rest. a. How much work does gravity do on the elevator
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Where does body heat come from?
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Friction produces heat

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