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ollegr [7]
3 years ago
7

Choose the +x-direction to point to the right. • Object 1 has a mass 1.66 kg and is moving to the right at 11.2 m/s. • Object 2

has a mass 6.59 kg and is moving to the left at 10.4 m/s. • What is x-component of the total momentum of the system in kg m/s? • Assume 3 significant figures for your final answer.
Physics
1 answer:
egoroff_w [7]3 years ago
5 0

Answer:

M = 49.4kgm/s (towards the left)

Explanation:

Momentum is the product of mass and velocity of an object

Momentum = mass * velocity

Momentum of Object 1 with mass 1.66 kg moving to the right at 11.2 m/s, is expressed as:

M1 = 1.66 * 11.2

M1 = 18.592kgm/s

Momentum of Object 2 with mass 6.59 kg moving to the left at 10.4 m/s is expressed as:

M2 = 6.59 * -10.4

M2 = -68.536kgm/s (negative since it is moving towards the left)

The total momentum will be the sum of momentum along the x-component as shown:

M = M1+M2

M = 18.592kgm/s--68.536kgm/s

M = -49.944kgm/s

M = 49.4kgm/s (towards the left)

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Say you want to make a sling by swinging a mass M of 1.9 kg in a horizontal circle of radius 0.042 m, using a string of length 0
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Answer: T= 715 N

Explanation:

The only external force (neglecting gravity) acting on the swinging mass, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:

T = mv² / r

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So the kinetic energy will be the following:

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\boxed{\large{\bold{\red{ANSWER~:) }}}}

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Airida [17]

Answer:

Complete question:

c.If the current in the second coil increases at a rate of 0.365 A/s , what is the magnitude of the induced emf in the first coil?

a.M= 6.53\times10^{-3} H

b.flux through each turn = Ф = 4.08\times10^{-4} Wb

c.magnitude of the induced emf in the first coil = e= 2.38\times10^{-3} V

Explanation:

a. rate of current changing = \frac{di}{dt}=[tex]M=\frac{1.60\times10^{-3} V}{0.240\frac{A}{s} }}[/tex]

  Induced emf in the coil =e= 1.60\times10^{-3} V

  For mutual inductance in which change in flux in one coil induces emf in the second coil given by the farmula based on farady law

     e=-M\frac{di}{dt}

     M=\frac{e}{\frac{di}{dt} }

     M=\frac{1.60\times10^{-3} }{-0.245}

   M= 6.53\times10^{-3} H

b.

  Flux through each turn=?

  Current in the first coil =1.25 A

   Number of turns = 20

       using   MI = NФ

     flux through each turn = Ф =  \frac{6.53\times10^{-3}\times1.25}{20}

   flux through each turn = Ф = 4.08\times10^{-4} Wb

c.

   second coil increase at a rate = 0.365 A/s

  magnitude of the induced emf in the first coil =?

 using   e=-M\frac{di_{2} }{dt}

            e= 6.53\times10^{-3} \times 0.365

magnitude of the induced emf in the first coil = e= 2.38\times10^{-3} V

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