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evablogger [386]
3 years ago
5

this is a non graded disscussion i want to fill out it is just a worksheet to help me get better 30 points

Physics
1 answer:
poizon [28]3 years ago
3 0

Answer:

1.) 30m/s

2) 5m/s

3) 40Hz

4) 400Hz

5) 300Hz

6) 0.77m

7) 0.386m

8) 0.625m

9) 100m/s

10) 50m/s

Explanation:

The speed of a wave (V) equals :

V = fλ ; where f = frequency, λ = wavelength

1.) f = 100Hz, λ = 0.30m

v = 100 * 0.30 = 30m/s

2) f = 50Hz λ = 0.10m

v = 50 × 0.10 = 5m/s

3) v = 20m/s, λ= 0.5m

f = v / λ

f = 20 / 0.5 = 40Hz

4) v = 80m/s, λ= 0.2m

f = 80/0.2 = 400Hz

5) v = 120m/s, λ = 0.4m

f = 120 / 0.4 = 300Hz

6) v = 340m/s, f = 440Hz

λ = v / f ; 340 / 440 = 0.77m

7) v = 340m/s, f = 880Hz

λ = 340 / 880 = 0.386m

8) v = 250m/s, f = 400

λ = 250 / 400 = 0.625m

9) f = 50Hz, λ = 2m

v = 50 × 2 = 100m/s

10) f = 100Hz, λ = 0.5m

v = 100 × 0.5 = 50m/s

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The sun emits __________ radiation, but it is radiated back off of planetary bodies as __________ radiation.
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Two blocks of clay, one of mass 100 kg and one of mass 3.00 kg, undergo a completely inelastic collision. Before the collision o
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Answer:

a)  v = 0.8 m / s , b)  v_{f} = 0.777 m / s , c) ΔK = 0.93 J

Explanation:

This exercise can be solved using the concepts of moment, first let's define the system as formed by the two blocks, so that the forces during the crash have been internal and the moment is conserved.

They give us the mass of block 1 (m1 = 100kg, its kinetic energy (K = 32 J), the mass of block 2 (m2 = 3.00 kg) and that it is at rest (v₀₂ = 0)

 

Before crash

     po = m1 vo1 + m2 vo2

     po = m1 vo1

After the crash

     p_{f} = (m1 + m2) v_{f}

a) The initial speed of the block of m1 = 100 kg, let's use the kinetic energy

     K = ½ m v²

     v = √2K / m

     v = √ (2 32/100)

     v = 0.8 m / s

b) The final speed,

    p₀ = p_{f}

    m1 v₀1 = (m1 + m2) v_{f}

   v_{f} = m1 / (m1 + m2) v₀₁

The initial velocity is calculated in the previous part v₀₁ = v = 0.8 m / s

    v_{f} = 100 / (3 + 100) 0.8

   v_{f} = 0.777 m / s

c) The change in kinetic energy

Initial      K₀ =K_{f}

              K₀ = 32 J

Final       K_{f} = ½ (m1 + m2) v_{f}²

              K_{f}= ½ (3 + 100) 0.777²

              K_{f} = 31.07 J

              ΔK = K_{f} - K₀

              ΔK = 31.07 - 32

              ΔK = -0.93 J

As it is a variation it could be given in absolute value

Part D

For this part s has the same initial kinetic energy K = 32 J, but it is block 2 (m2 = 3.00kg) in which it moves

d) we use kinetic energy

        v = √ 2K / m2

        v = √ (2 32/3)

        v = 4.62 m / s

e) the final speed

      v₀₂ = v =  4.62 m/s  

      p₀ = m2 v₀₂

      m2 v₀₂ = (m1 + m2) v_{f}

      v_{f} = m2 / (m1 + m2) v₀₂

      v_{f} = 3 / (100 + 3) 4.62

      v_{f} = 0.135 m / s

f) variation of kinetic energy

     K_{f} = ½ (m1 + m2) v_{f}²

     K_{f} = ½ (3 + 100) 0.135²

     K_{f} = 0.9286 J

     ΔK = 0.9286-32

    ΔK = 31.06 J

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                                         F=G \frac{M \times m}{r^{2}}

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F – Force, G – gravitational constant, M and m – masses in kg, r – distance in meters.

Since force is proportional to the masses of interacting objects. If the mass of any one object increases, gravity between them also gets increased. When moving to higher altitude, force decreases as the distance is inverse proportion to gravity.

4 0
3 years ago
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