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lisov135 [29]
2 years ago
8

Two students, sitting on frictionless carts, push against each other. Both are initially at rest and the mass of student 1 and t

he cart is M, and that of student 2 and the cart is 1.5M. If student 1 pushes student 2 so that she recoils with velocity v,  what is the velocity of student 2.
Physics
1 answer:
Zepler [3.9K]2 years ago
6 0

Answer:

  v₂ = v/1.5= 0.667 v

Explanation:

For this exercise we will use the conservation of the moment, for this we will define a system formed by the two students and the cars, for this isolated system the forces during the contact are internal, therefore the moment conserves.

Initial moment before pushing

    p₀ = 0

Final moment after they have been pushed

    p_{f} = m₁ v₁ + m₂ v₂

   p₀ =  p_{f}

   0 = m₁ v₁ + m₂ v₂

   m₁ v₁ = - m₂ v₂

Let's replace

   M (-v) = -1.5M v₂

   v₂ = v / 1.5

  v₂ = 0.667 v

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Answer:

Speed of the wave is 7.87 m/s.

Explanation:

It is given that, tapping the surface of a pan of water generates 17.5 waves per second.

We know that the number of waves per second is called the frequency of a wave.

So, f = 17.5 Hz

Wavelength of each wave, \lambda=45\ cm=0.45\ m

Speed of the wave is given by :

v=f\lambda

v=17.5\times 0.45

v = 7.87 m/s

So, the speed of the wave is 7.87 m/s. Hence, this is the required solution.

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2 years ago
The air in a pipe resonates at 150 Hz and 750 Hz, one of these resonances being the fundamental. If the pipe is open at both end
Xelga [282]

Answer:

Explanation:

Two frequencies with magnitude 150 Hz and 750 Hz are given

For Pipe open at both sides

fundamental frequency is 150 Hz as it is smaller

frequency  of pipe is given by

f=\frac{nv}{2L}

where L=length of Pipe

v=velocity of sound

f=150\ Hz for n=1

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thus there are three resonance frequencies for n=2,3 and 4

For Pipe closed at one end

frequency is given by

f=\frac{(2n+1)}{4L}\cdot v

for n=0

f_1=\frac{v}{4L}

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2 years ago
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boyakko [2]

Answer:

final displacement = +24484.5 nm

Explanation:

The path difference when 158 bright spots were observed with red light (λ1 = 656.3 nm) is given as;

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So, 2d2 - 2d1 = 150λ1

Dividing both sides by 2 to get;

d2 - d1 = 75λ1 - - - - eq1

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d1 = distance between the fixed mirror and the beam splitter

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Now, the path difference between the two waves when 114 bright spots were observed is;

Δr = 2d'2 - 2d1 = 114λ1

2d'2 - 2d1 = 114λ1

Divide both sides by 2 to get;

d'2 - d1 = 57λ1

Where;

d'2 is the new position of the movable mirror from the splitter

Now, the displacement of the moveable mirror is (d2 - d'2). To get this, we will subtract eq2 from eq1.

(d2 - d1) - (d'2 - d1) = 75λ1 - 57λ2

d2 - d1 - d'2 + d1 = 75λ1 - 57λ2

d2 - d'2 = 75λ1 - 57λ2

We are given;

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Thus;

d2 - d'2 = 75(656.3) - 57(434)

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Answer:

Explanation:

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<u><em /></u>

<u><em>Got The Answer From Google</em></u>

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