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Thepotemich [5.8K]
3 years ago
15

A 480 kg car moving at 14.4 m/s hits from behind another car moving at 13.3 m/s in the same direction. If the second car has a m

ass of 570 kg and a new speed of 17.9 m/s, what is the velocity of the first car after the collision?
Physics
1 answer:
Nostrana [21]3 years ago
5 0

Answer:

Velocity of the first car after the collision, v_1=8.93\ m/s

Explanation:

It is given that,

Mass of the car, m_1 = 480\ kg

Initial speed of the car, u_1 = 14.4\ m/s    

Mass of another car, m_2 = 570\ kg

Initial speed of the second car, u_2 = 13.3\ m/s  

New speed of the second car, v_2 = 17.9\ m/s  

Let v_1 is the final speed of the first car after the collision. The total momentum of the system remains conserved, Using the conservation of momentum to find it as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

m_1u_1+m_2u_2-m_2v_2=m_1v_1

480\times 14.4+570\times 13.3-570\times 17.9=480v_1

v_1=8.93\ m/s

So, the velocity of the first car after the collision is 8.93 m/s. Hence, this is the required solution.

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What force is necessary to accelerate a 2500 kg care from rest to 20 m/s over 10 seconds?
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What is Tension variables?
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The tension on an object is equal to the mass of the object x gravitational force plus/minus the mass x acceleration. T = mg + ma.

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3 years ago
A standing wave is created on a string of length 1.2 m. If the speed of the wave on the string is 60.0 m/s, what is the fundamen
coldgirl [10]

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Fundamental frequency in the string will be 25 Hz

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Read 2 more answers
Suppose that the height of the incline is h = 14.7 m. Find the speed at the bottom for each of the following objects. 1.solid sp
tensa zangetsu [6.8K]

Answer:

1. 14.4 m/s  2. 13.2 m/s 3. 12.0 m/s 4. 13.9 m/s

Explanation:  

Assuming no friction present, the different objects roll without slipping, so there is a constant relationship between linear and angular velocity, as follows:

ω= v/r

If no friction exists, the change in total kinetic energy must be equal in magnitude to the change in the gravitational potential energy:

∆K = -∆U

 ½ *m*v² + ½* I* ω²  = m*g*h

Simplifying and replacing the value of the angular velocity:

½ * v² + ½ I *(v/r)² = g*h (1)

In order to answer the question, we just need to replace h by the value given, and I (moment of inertia) for the value for each different object, as follows:

  •  Solid Sphere I = 2/5* m *r²

                Replacing in (1):

                ½ * v² + ½ (2/5 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(10/7*9.8 m/s2*14.7 m)  = 14. 4 m/s

  • Spherical shell I=2/3*m*r²

                Replacing in (1):

                ½ * v² + ½ (2/3 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(6/5*9.8 m/s2*14.7 m)  = 13.2 m/s

  • Hoop   I= m*r²

                Replacing in (1):

                ½ * v² + ½ (m*r²) *(v/r)² = g*h

               Replacing by the value given for h, and solving for v:

               v = √(9.8 m/s2*14.7 m)  = 12.0 m/s

  • Cylinder I = 1/2 * m* r²

                 Replacing in (1):

                ½ * v² + ½ (1/2 *m*r²) *(v/r)²= g*h

                 Replacing by the value given for h, and solving for v:

                v = 2*√(1/3*9.8 m/s2*14.7 m)  = 13.9 m/s

5 0
3 years ago
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