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JulsSmile [24]
3 years ago
8

If 2 objects had the same momentum, what must be true about the mass of the object that traveled the fastest?

Physics
1 answer:
julsineya [31]3 years ago
4 0

Yes, the above-given statement is true

<u>Explanation:</u>

  • The product of the mass x the velocity will be the same for both. Momentum is the action of a body with a particular mass through space and there is the conservation of momentum.
  • Momentum is described as the mass of the object multiplied by its velocity.
  • <u>Momentum (p) = Mass (M) * Velocity (v)</u>
  • Therefore for two objects with many masses to have a similar momentum, then the lighter one has to be moving quicker than the heavier object.

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Answer:

C. The left wire attracts the right wire and exerts as much force as the right wire does.

Explanation:

To know what is the answer you first take into account the magnetic field generated by each current, for a distance of d:

B_1=\frac{\mu_oI_1}{2\pi d}=\frac{\mu_o}{2\pi d}(1A)\\\\B_2=\frac{\mu_oI_2}{2\pi d}=\frac{\mu_o}{2\pi d}(2A)=2B_1\\\\

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\vec{F_B}=I\vec{L}\ X \vec{B}

if the direction of the L vector is in +k direction, the first wire produced a magnetic field with direction +y, that is, +j and the second wire produced magnetic field with direction -y, that is, -j (this because the direction of the magnetic field is obtained by suing the right hand rule). Hence, the direction of the magnetic force on each wire, produced by the other one is:

\vec{F_{B1}}=I_1L\hat{k}\ X\ B_2(-\hat{j})=I_1LB_2\hat{i}=(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}\\\\\vec{F_{B2}}=I_2L\hat{k}\ X\ B_2(\hat{j})=I_2LB_1\hat{i}=-(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}

Hence, due to this result you have that:

C. The left wire attracts the right wire and exerts as much force as the right wire does.

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