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enot [183]
4 years ago
14

What is the name of a length that is 100 times larger than a millimeter

Physics
1 answer:
Pie4 years ago
4 0
<span>The scales is 1 m = 10 dm = 100 cm = 1000 mm. Then 10 dm equals 1000 mm and 1 dm equals 100 mm.The names of the measures in that scale are meter, decimeter, centimeter and milimiter. So, the answer is that the length that is 100 times larger than a milimieter is the decimeter (dm). </span>
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Answer:

Yes

Explanation:

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3 years ago
If eight water waves pass an ocean buoy each minute, and successive wave crests are 20 m apart, find the wave speed:____________
Llana [10]

Answer:

The wave speed is calculated below:

Explanation:

Given,

number of waves passed per minute = 8

time period = 1 minute = 60 s

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waves passing interval per second = \frac{8}{60} s^{-1}

Now,

wave speed = 20 m × \frac{8}{60} s^{-1}

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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)

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      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}

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These will transform internal system energy

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      v₂₀ ‘= v₂₀ -u

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      v_{f} ‘=   v_{f} - u

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Initial

    K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²

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    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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