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VikaD [51]
3 years ago
12

What do low energy waves have

Physics
1 answer:
Elena-2011 [213]3 years ago
7 0
The Answer: Photons

There is photons in low energy waves
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A certain type of bird lives in two regions of a state. The distribution of weight for birds of this type in the northern region
Annette [7]

Answer: (a) Z-score are 1 and -1.2 for northern and southern regions, respectively.

Explanation: <u>Z-score</u> is how many standard deviations a data is from the population mean or how far a data point is from the mean.

The z-score is calculated by the following:

z=\frac{x-\mu}{\sigma}

where

x is the data point

μ is population mean

σ is standard deviation

For the <u>northern</u> <u>region</u> birds:

μ = 10, σ = 3, x = 13

z=\frac{13-10}{3}

z = 1

The z-score for birds living in the northern region is 1, which means it is 1 standard deviation <em>above the mean</em>.

For the southern region:

μ = 16, σ = 2.5, x = 13

z=\frac{13-16}{2.5}

z = -1.2

The z-score for southern living birds is -1.2, meaning it is 1.2 standard deviations <em>below the mean</em>.

6 0
3 years ago
You toss a 0.40-kg ball at 9.0 ms/ to a 14-kg dog standing on an iced-over pond. The dog catches the ball and begins to slide on
just olya [345]

Answer:

a)   v_{f} = 0.25 m / s  b) u = 0.25 m / s

Explanation:

a) To solve this problem let's start with the conservation of the moment, for this we define a system formed by the ball plus the dog, in this case all the forces are internal and the moment is conserved

We will write the data

     m₁ = 0.40 kg

     v₁₀ = 9.0 m / s

     m₂ = 14 kg

     v₂₀ = 0

Initial

     po = m₁ v₁₀

Final

     p_{f} = (m₁ + m₂) vf

     po = pf

     m₁ v₁₀ = (m₁ + m₂) v_{f}

      v_{f} = v₁₀ m₁ / (m₁ + m₂)

      v_{f} = 9.0 (0.40 / (0.40 +14)

      v_{f} = 0.25 m / s

b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass

      u = 0.25 m / s

In the direction of movement of the ball

c) Let's calculate the kinetic energy in both moments

Initial

     K₀ = ½ m₁ v₁₀² +0

     K₀ = ½ 0.40 9 2

     K₀ = 16.2 J

Final

     K_{f}= ½ (m₁ + m₂) v_{f}2

      K_{f} = ½ (0.4 +14) 0.25 2

    K_{f} = 0.45 J

   

    ΔK = K₀ -  K_{f}

    ΔK = 16.2-0.445

    ΔK = 1575 J

These will transform internal system energy

d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.

      v₁₀’= v₁₀ -u

      v₁₀’= 9 -.025

      v₁₀‘= 8.75 m / s

      v₂₀ ‘= v₂₀ -u

      v₂₀‘= - 0.25 m / s

      v_{f} ‘=   v_{f} - u

      v_{f} = 0

Initial

    K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²

    Ko = ½ 0.4 8.75² + ½ 14.0 0.25²

    Ko = 15.31 + 0.4375

    K o = 15.75 J

Final

   k_{f} = ½ (m₁ + m₂) vf’²

  k_{f} = 0

All initial kinetic energy is transformed into internal energy in this reference system

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