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Alik [6]
3 years ago
11

Scientific knowledge is generally considered to have a positive effect on society. In which situation would scientific knowledge

most likely have a negative effect on society?
Physics
1 answer:
attashe74 [19]3 years ago
5 0
Scientific knowledge itself cannot have a positive or negative impact on society per se. What can have a negative or positive impact is how this knowledge is being used. 

While we all enjoy a beter life due to advancements in medicine and technology for example, this was because people decided they would put certain scientific discoveries into practice and spread them among the people.

The same thing can also be said for negative scientific discoveries. While energy produced in atomic power plants per se isn't harmful, when making an atomic bomb out of it, it can be disastrous. It was the intention of doing something evil with it that might have had a negative impact on society, not the knowledge itself. 
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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
3 years ago
Two tuning forks of frequency 480 hz and 484 hz are struck simultaneously. what is the beat frequency resulting from the two sou
umka21 [38]
Ans: Beat frequency = f_b = 4Hz

Explanation: 
The beat frequency is equal to the absolute value of the difference in frequency of the two waves. In other words, the number of beats per second is equal to the difference in frequency. It is due to the destructive and constructive interference. <span>According to this interference, sound will be soft or loud.

Hence. the formula is:
</span>Beat frequency = f_b = |f_2 - f_1|
<span>
Since,
</span>f_1 = 480Hz
f_2 = 484Hz

Therefore,
Beat frequency = f_b = |484 - 480|

=> Beat frequency = f_b = 4Hz
-i
7 0
3 years ago
Hbkvljgvnggvkngvljhvlmhvglhbkjghv
Schach [20]
Jaydbrbstgsbwiybsbdd to nebsndhbd
4 0
3 years ago
Derive the formula v= u+ at<br>please derive in detail ​
gogolik [260]

Lets do

We know

The rate of change of velocity is acceleration .

\\ \sf\longmapsto a=\dfrac{dv}{dt}

\\ \sf\longmapsto dv=adt

Integrate both sides

\\ \sf\longmapsto \int dv=a\int dt

As acceleration is constant .Take it outside of integral .On velocity we can take limit u to v and time from 0 to t

\\ \sf\longmapsto {\displaystyle{\int}}^v_u dv=a{\displaystyle{\int}}^t_0 dt

Hence

\\ \sf\longmapsto v{\huge{|}}^v_u=at

\\ \sf\longmapsto v-u=at

\\ \sf\longmapsto v=u+at

4 0
3 years ago
Read 2 more answers
The period of an ocean wave is 5 seconds. What is the wave's frequency?
Oksana_A [137]

Answer:

f=1/5= 0.2 Hz

..,...........

7 0
3 years ago
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