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zhuklara [117]
3 years ago
12

Just wondering if I did this right

Physics
1 answer:
Kisachek [45]3 years ago
5 0

Yeah

All they are all correct

You might be interested in
During a circus performance, a 72-kg humancannonball is shot out of an 18-m-long cannon. If thehuman cannonball spends 0.95 s in
andreyandreev [35.5K]

Answer:

2872.8 N

Explanation:

We have the following information

m =n72kg

Δy = 18m

t = 0.95s.

From here we use the equation

Δy=1/2at2 in order to solve for the acceleration.

So a

=( 2x 18m)/(0.95s²)

= 36/0.9025

= 39.9m/s2.

From there we use the equation

F = ma

F=(72kg) x (39.9)

= 2872.8N.

2872.8N is the average net force exerted on him in the barrel of the cannon.

Thank you!

7 0
3 years ago
Bruh how did i get 3 warnings and 3 brainliests removed
Brums [2.3K]

Answer:

Wait, that can happen? I'm sorry.

Explanation:

5 0
3 years ago
Read 2 more answers
A 55 kg person falling with a velocity of 0.6
siniylev [52]

Answer:

What are we supposed to find, if it is kinetic energy then this is the solution.

K.E=1/2mv^2

K.E= kinetic energy

M=mass

V=velocity

K.E =0.5*55*0.6^2

K.E=9.9J

Explanation:

3 0
3 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
3 years ago
Johnson made a hole at the bottom of a plastic bottle containing water. However, he noticed that the water did not flow out from
Sphinxa [80]

Answer:

  • a) See explanation below

  • b) At X.

Explanation:

Please, see the picture attached with the image of the plastic bottle for this question.

<u>(a) Explain why the water could not flow out of the bottle.</u>

What makes the water flow out of the botlle is the force of gravity, whic attracts the water towards the Earth.

When Johnson made a small hole at the bottom of the plastic bottle containing water, the air outside the bottle, which surrounds it and exerts a pressure all over the outer walls of the bottle, exerted a force against the small area of water "over" the hole that is in contact with the air.

Thus, this force of the air pushing upward through the wall opposed the force of gravity pulling downward making the net force zero and the water cannot fall.

<u>(b) To make the water flow out more easily, his teacher suggested making another hole. At which position - X, Y or Z, should he make the 2nd hole in order for the water to flow out the fastest?</u>

You must open the hole at a place where there is not water but air, such that the outer air can enter in the bottle.

That will make that the pressure in the space over the water inside the bottle be equal to the pressure outside.

The pressure of the air above the water will push it downward. Now, the force from the pressure of air inside the water, which is downward, opposes the upward force from the pressure of air around the first hole, and the net force will be downward, making the water flow out more easily.

Thus, the position where he should make the second hole in order for the water to flow fastest is at X.

4 0
4 years ago
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