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kipiarov [429]
3 years ago
10

Identified __________ should be tracked until they are corrected

Physics
1 answer:
Reika [66]3 years ago
3 0
<h3>✽ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ✽</h3>

➷ It would be A. Hazards

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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Which of the following best describes the difference between type a and type b
Harrizon [31]
Is there supposed to be a picture? Next time try putting a picture
3 0
3 years ago
Cindy runs 2 kilometers every morning. She takes 2 minutes for the first 250 meters, 4 minutes for the next 1,000 meters, 1 minu
PIT_PIT [208]

Answer:

200 m / min

Explanation:

Total distance = 2000 m

Total time = 2+4+1+3 = 10 minutes

Average speed = 2000 m / 10 min = 200 m/min

7 0
1 year ago
A Block slides down an incline that makes an angle of 30? with the horizontal direction. The coefficient of kinetic friction bet
astraxan [27]

Answer:

Acceleration = 2.35 m/s^{2}

Speed = 8.67 m/s

Explanation:

The coefficient of friction , u =0.3

The angle of incline = 30°

The two forces acting on block are weight and friction.

weight along the incline = mg cos60° = \frac{mg}{2} = 0.5 mg

Friction along incline = umg cos30° = mg 0.3\times \frac{\sqrt{3}}{2}

Friction along incline  = 0.26 mg

Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg

Acceleration = \frac{net force}{mass} = 0.24 g = 2.35 m/s^{2}

The height of incline = 8 m

Length of the inclined edge = 16 m

v^{2}=u^{2}+2as

v^{2}= 2\times 0.24 \times 9.8\times 16

v= 8.67 m/s

5 0
3 years ago
Use the graph below to answer the following question: if average acceleration is calculated using the equation, “ change in velo
sergiy2304 [10]

Answer:

a=9\ cm/s^2

Explanation:

<u>Average Acceleration </u>

Acceleration is a physical magnitude defined as the change of velocity over time. When we have experimental data, we can compute it by calculating the slope of the line in velocity vs time graph.

Note: <em>We cannot see if the time axis is numbered in increments of 1 second, and we'll assume that. </em>

When t_2=4\ sec, the graph shows a value of v_2=36\ cm/s

When t_1=0\ sec, the object is at rest, v_1=0

We compute the average acceleration as

\displaystyle a=\frac{v_2-v_1}{t_2-t_1}

\displaystyle a=\frac{36\ cm/s-0\ cm/s}{4\ sec-0\ sec}

\displaystyle a=\frac{36\ cm/s}{4\ s}

\boxed{a=9\ cm/s^2}

6 0
2 years ago
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
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