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Yuki888 [10]
4 years ago
11

What does a mu mean in physics

Physics
1 answer:
ivolga24 [154]4 years ago
3 0

'mu' is a letter in the Greek alphabet.  It looks like this:  μ
If you need labels for quantities or variables in math or science,
you can use any letter from any alphabet you like ... just be sure
to explain what the symbol stands for somewhere on the page
where the equation or formula is.

I've seen 'mu' used to stand for 'micro-' when writing units.
For example:

" μ-meter " ... one millionth of a meter

" μ-farad " ... one millionth of a farad

" μ-second " ... one millionth of a second






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Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1200 kg and was approaching at
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Answer:

a) v = 11.24 m / s ,    θ = 17.76º   b) Kf / K₀ = 0.4380

Explanation:

a) This is an exercise in collisions, therefore the conservation of the moment must be used

Let's define the system as formed by the two cars, therefore the forces during the crash are internal and the moment is conserved

Recall that moment is a vector quantity so it must be kept on each axis

X axis

initial moment. Before the crash

     p₀ₓ = m₁ v₁

where v₁ = -25.00 me / s

the negative sign is because it is moving west and m₁ = 900 kg

final moment. After the crash

      p_{x f}= (m₁ + m₂) vx

       p₀ₓ =  p_{x f}

       m₁ v₁ = (m₁ + m₂) vₓ

     vₓ = m1 / (m₁ + m₂) v₁

let's calculate

       vₓ = - 900 / (900 + 1200) 25

       vₓ = - 10.7 m / s

Axis y

initial moment

      p_{oy}= m₂ v₂

where v₂ = - 6.00 m / s

the sign indicates that it is moving to the South

final moment

     p_{fy}= (m₁ + m₂) v_{y}

     p_{oy} = p_{fy}

     m₂ v₂ = (m₁ + m₂) v_{y}

     v_{y} = m₂ / (m₁ + m₂) v₂

we calculate

    v_{y} = 1200 / (900+ 1200) 6

    v_{y}  = - 3,428 m / s

for the velocity module we use the Pythagorean theorem

      v = √ (vₓ² + v_{y}²)

      v = RA (10.7²2 + 3,428²2)

      v = 11.24 m / s

now let's use trigonometry to encode the angle measured in the west clockwise (negative of the x axis)

      tan θ = v_{y} / Vₓ

      θ = tan-1 v_{y} / vₓ)

      θ = tan -1 (3,428 / 10.7)

       θ = 17.76º

This angle is from the west to the south, that is, in the third quadrant.

b) To search for loss of the kinetic flow, calculate the kinetic enegy and then look for its relationship

      Kf = 1/2 (m1 + m2) v2

      K₀ = ½ m₁ v₁² + ½ m₂ v₂²

      Kf = ½ (900 + 1200) 11.24 2

      Kf = 1.3265 105 J

      K₀ = ½ 900 25²  + ½ 1200 6²

      K₀ = 2,8125 10⁵ + 2,16 10₅4

        K₀ = 3.0285 105J

the wasted energy is

        Kf / K₀ = 1.3265 105 / 3.0285 105

        Kf / K₀ = 0.4380

         

this is the fraction of kinetic energy that is conserved, transforming heat and transforming potential energy

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