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xeze [42]
3 years ago
14

Object A has of mass 7.20 kilograms, and object B has a mass of 5.75 kilograms. The two objects move along a straight line towar

d each other with velocities +2.00 meters/second and -1.30 meters/second respectively. What is the total kinetic energy of the objects after the collision, if the collision is perfectly elastic?
Physics
2 answers:
cestrela7 [59]3 years ago
8 0
Two things:
1) All collisions conserve momentum
2) Elastic collisions also conserve kinetic energy.

So basically the kinetic energy before the collision is the same as the kinetic energy after the collision.

KE = ½mv²

For object A:
KE = ½(7.20 kg)(2.00 m/s)² = 14.4 J

For object B:
KE = ½(5.75 kg)(-1.30 m/s)² = 4.86 J

total kinetic energy = 14.4 J + 4.86 J = 19.3 J
Genrish500 [490]3 years ago
7 0
The equation for Kinetic Energy in an Elastic Collision is this:
1/2(m₁v₁) + 1/2(m₂v₂) = 1/2(7.20kg)(2.00m/s) + 1/2(5.75kg)(-1.30m/s) =
3.4625m/s

I think....technically the equation above equates to the same thing except measuring the final velocity of the given masses, however you were not given that in the equation, making the first half of the full equation the only viable solution.

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Playing a certain musical note on a trumpet
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(1) Resonance

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2 years ago
horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 3 kilograms is tied to the middle of the clothes
lbvjy [14]

Answer:

The tension is  T =  103.96N

Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

           The distance between the two poles is D =14 m

          The mass tied between the two cloth line is  m = 3Kg

         The distance it sags is d_s = 1m

The objective of this solution is to obtain the magnitude of the tension on the ends of the  clothesline

Now the sum of the forces on the y-axis is zero assuming  that the whole system is at equilibrium

       And this can be mathematically represented as

                             \sum F_y = 0

 To obtain \theta we apply SOHCAHTOH Rule

 So    Tan \theta = \frac{opp}{adj}

          \theta = tan^{-1} [\frac{opp}{adj} ]

            = tan^{-1} [\frac{1}{7}]

          =8.130^o

=>  \  \ \ \ \ \ \ \ 2T sin\theta -mg =0

=>  \  \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}

=>  \  \ \ \ \ \ \ \ T = \frac{3 * 9.8 }{2 sin \theta }

=>  \  \ \ \ \ \ \ \  T =\frac{29.4}{2sin(8.130)}

=>  \  \ \ \ \ \ \ \  T = 103.96N

             

                 

5 0
2 years ago
Read 2 more answers
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