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OLga [1]
3 years ago
15

What is a non example of temperature

Physics
2 answers:
almond37 [142]3 years ago
7 0

We measure temperature of matters with a device called thermometer. There are three types of thermometer, Celsius Thermometer, Fahrenheit Thermometer and Kelvin Thermometer.

Nimfa-mama [501]3 years ago
5 0

Best Answer

1) what do you mean "non -examples"

2) energy is something in the physical world. we cannot see fell hear touch etc

but it had different forms: work = pushing or pulling a distance, lifting, moving, heat, light sound electricity, . it is best understood by giving examples of the different forms.

it is an idea that is only about 400 years old in physics

It does NOT mean excitement, as "the football crowd had a lot of energy after the big game"

I like to say it is what ever it is, that allows physical change to happen

3) temperature is what we feel as hot or cold and is a measure of average heat energy of a material.

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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
A 7.5-cmcm-diameter horizontal pipe gradually narrows to 4.5 cmcm . When water flows through this pipe at a certain rate, the ga
tino4ka555 [31]

Answer

given,

diameter,d₁ = 7.5 cm

               d₂ = 4.5 cm

P₁ = 32 kPa

P₂ = 25 kPa

Assuming, we have calculation of flow in the pipe

using continuity equation

 A₁ v₁ = A₂ v₂

 π r₁² v₁ = π r₂² v₂

 v_1= \dfrac{r_2^2}{r_1^2} v_2

 v_1= \dfrac{2.25^2}{3.75^2} v_2

 v_1= 0.36 v_2

Applying Bernoulli's equation

 \Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)

 P_1-P_2 = \dfrac{1}{2}\rho (v_2^2-(0.36 v_2)^2)

 32-25 = \dfrac{1}{2}1000\times v_2^2 (1 - 0.1269)

 v_2=\sqrt{\dfrac{2\times 7\times 10^3}{1000\times (0.8704)}}

 v_2=\sqrt{16.084}

       v₂ = 4.01 m/s

fluid flow rate

Q = A₂ V₂

Q = π (0.0225)²  x 4.01

Q = 6.38 x 10⁻³ m³/s

flow in the pipe is equal to 6.38 x 10⁻³ m³/s

4 0
3 years ago
What is lenz law?<br>What is galvanometer?​
yulyashka [42]

Answer:

Lenz's law, in electromagnetism, statement that an induced electric current flows in a direction such that the current opposes the change that induced it. This law was deduced in 1834 by the Russian physicist Heinrich Friedrich Emil Lenz (1804–65).

6 0
2 years ago
A 20-Kg child is on a swing attached to 3.0 m-long chains. The child swings back and forth, swinging out to a 60-degree angle. (
kvv77 [185]

Answer:

 v = 29.4 m / s

Explanation:

For this exercise we can use the conservation of mechanical energy

Lowest starting point.

          Em₀ = K = ½ m v²

final point. Higher

          Em_{f} = U = m g h

Let's use trigonometry to lock her up

          cos 60 = y / L

          y = L cos 60

Height is the initial length minus the length at the maximum angle

           h = L - L cos 60

           h = L (1- cos 60)

energy is conserved

         Em₀ = Em_{f}

          ½ m v² = mgL (1 - cos 60)

         v = 2g L (1- cos 60)

 

let's calculate

          v² = 2 9.8 3.0 (1- cos 60)

          v = 29.4 m / s

6 0
3 years ago
Which scenario involves kinetic energy transforming into potential energy? A. a book lying on a shelf B. a train traveling at a
NeX [460]
The answer is C.

The Kinetic energy which was exerted and experience pulling the string of a bow is kept as a potential energy at the end of the arrow in contact with the string. Once release from aim at stationary position the potential energy is again transformed.
6 0
4 years ago
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