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Elden [556K]
3 years ago
8

Two objects, X and Y, are held at rest on a horizontal frictionless surface and a spring is compressed between them. The mass of

X is 2/5 times the mass of Y. Immediately after the spring is released, X has a kinetic energy of 50 J and Y has a kinetic energy of:
Physics
1 answer:
Ghella [55]3 years ago
7 0

Answer:

20J

Explanation:

Using conservation law of momentum;

since the bodies were at rest, their initial momentum is zero

0 =  M1Vx + M2Vy

- M1Vx = M2Vy where Vx is the final velocity of x after the spring has been release and Vy is final velocity of y and M1 and M2 are the masses of x and y

also M1 = 2/5 M2

substitute M1 into the the equation above

-2/5 M2Vx =  M2Vy

cancel M2 on both side

-2/5Vx =  Vy

comparing the kinetic energy of both x and y

for x K.E = 1/2 M1 Vx²

and y K.E = 1/2M2 Vy²

substitute for M1 = 2/5 M2

K.Ex = 1/2 × 2/5 M2 Vx²

divide  K.Ex / K.Ey = (1/2 × 2/5 M2 Vx²) / 1/2 M2 Vy²

cancel the common terms

K.Ex / K.Ey = (2/5 Vx²) / Vy²

substitute -2/5Vx for Vy

(2/5 Vx²) / ( -2/5 Vx)² = (2/5 Vx²) / ( 4/25 Vx²)

cancel Vx²

(2/5) / (4/25) = 2/5 ÷ 4/25 = 2/5 × 25/4 = 5/2

the ratio of x and y kinetic energy is 5:2

since the kinetic energy of x is 50

50 : 20 = 5 : 2 if 10 is used to divide both sides

the kinetic energy of y = 20 J

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The viewing screen in a double-slit experiment with monochromatic light. Fringe C is the central maximum. The fringe separation
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Answer:

<em>Part A</em><em>:</em>

a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.

<em>Part B</em><em>:</em>

b) If the spacing between the slits is decreased the fringe spacing Δy will increase.

<em>Part C</em><em>:</em>

a) If the distance to the screen is decreased the fringe spacing will decrease.

<em>Part D</em><em>:</em>

The dot in the center of fringe E is 920\ x\ 10^{-9} m farther from the left slit than from the right slit.

Explanation:

In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.

The position of bright fringes in the screen where the pattern is formed can be calculated with

                      \vartriangle y =\frac{m \lambda L}{d}

                      m=0,\pm 1,\pm 2,\pm 3,.....

  1. m is the order number.
  2. \lambda is the wavelength of the monochromatic light.
  3. L is the distance between the screen and the two slits.
  4. d is the distance between the slits.
  • Part A:  a) In the above equation for the position of bright fringes we can see that if the wavelength of the light \lambda is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.
  • Part B:  b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
  • Part C:  a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.
  • Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at  the center of the fringe E in the screen we use the condition for constructive interference. That says that the  path length difference Δr between rays coming from the left and right slit must be \vartriangle r=m \lambda

        We simply replace the values in that equation :

                      \vartriangle r= m \lambda =2.\ 460\ nm

                      \vartriangle r= 920\ x\ 10^{-9} m

         The dot in the center of fringe E is 920\ x\ 10^{-9}m farther from the left slit than from the right slit.

     

       

       

     

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Two skaters, a man and a woman, are standing on ice. Neglect any friction between the skate blades and the ice. The mass of the
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Answer:

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a₂ = -1.18 m/s²  (West)

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a₁ = ?

a₂ = ?

To get the acceleration (magnitude and direction) of the man we apply

∑Fx = m*a   (⇒)

F = m₁*a₁         ⇒      60 N = 95 Kg*a₁    

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∑Fx = m*a   (⇒)

F = -m₂*a₂         ⇒      60 N = -51 Kg*a₂    

⇒  a₂ = (60N / 51Kg) = -1.18 m/s²  (West)

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