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ira [324]
2 years ago
8

Not yet answered Points out of 2.00 Not flaggedFlag question Question text Two cars collide head-on on a level friction-free roa

d. The collision was completely inelastic and both cars quickly came to rest during the collision. What is true about the velocity of this system's center of mass?
Physics
1 answer:
Natali5045456 [20]2 years ago
5 0

Answer:

Velocity of center of mass must be ZERO

Explanation:

As we know that two cars are moving on friction-less surface

So here we can say that on the system of cars there is no external force along x direction

So we can say that

F_{net} = \frac{dP_{cm}}{dt}

so here if net force is zero

\frac{dP_{cm}}{dt} = 0

P_{cm} = constant

since finally the two cars comes to rest so here final velocity of center of mass of two cars will be zero and hence it will be always zero as there is no force on the system of two cars

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With a mass of 109 kg, Baby Bird is the smallest monoplane ever
OLga [1]

Total resultant velocity=5.11-3.27=1.84m/s

  • m_1=61.4kg
  • m_2=109kg
  • v_1=1.84m/s
  • v_2=?

\\ \sf\longmapsto ∆P=P

\\ \sf\longmapsto m_1v_1=m_2v_2

\\ \sf\longmapsto v_2=\dfrac{m_1v_1}{m_2}

\\ \sf\longmapsto v_2=\dfrac{61.4(1.84)}{109}

\\ \sf\longmapsto v_2=112.976/109

\\ \sf\longmapsto v_2\approx 1.3m/s

4 0
3 years ago
9.58 A spring of equilibrium length L1 and spring constant k1 hangs from the ceiling. Mass m1 is suspended from its lower end. T
andrey2020 [161]

Answer:

The distance of m2 from the ceiling is L1 +L2 + m1g/k1 + m2g/k1 + m2g/k2.

See attachment below for full solution

Explanation:

This is so because the the attached mass m1 on the spring causes the first spring to stretch by a distance of m1g/k1 (hookes law). This plus the equilibrium lengtb of the spring gives the position of the mass m1 from the ceiling. The second mass mass m2 causes both springs 1 and 2 to stretch by an amout proportional to its weight just like above. The respective stretchings are m2g/k1 for spring 1 and m2g/k2 for spring 2. These plus the position of m1 and the equilibrium length of spring 2 L2 gives the distance of L2 from the ceiling.

4 0
3 years ago
A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are
igor_vitrenko [27]

Answer:

0.231 N

Explanation:

To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be

\alpha = \frac{\theta}{t} = \frac{6.37}{9.87} = 0.6454 rad/s^2

If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:

I = \frac{mL^2}{3} = \frac{1.27*0.847^2}{3} = 0.303kgm^2

According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is

T = I\alpha = 0.303*0.6454 = 0.196 Nm

So the force acting on the other end to generate this torque mush be:

F = \frac{T}{L} = \frac{0.196}{0.847} = 0.231 N

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3 years ago
Como un astronauta utiliza la fuerza de hallar y empujar?
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Answer:

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Explanation:

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coldgirl [10]

Answer:

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