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Zigmanuir [339]
2 years ago
5

jamal is playing with magnetic toy vehicles. He has two identical magnetic vehicles (purple and green) on different sides of a c

enter magnet that cannot move. His friend Simone challenges him to move one vehicle one space to the left or the right in order to get the largest increase in potential energy. That means Jamal can move the purple car to point A or point B, or he can move the green car to point C or point D. Explain which movement Jamal should make (which car and which point), and why that movement will result in the largest increase in potential energy. Describe the magnetic force that will act on the vehicle he moves.
Physics
1 answer:
KengaRu [80]2 years ago
5 0

Answer:

Since potential energy increases with increase in the separation distance, Jamal should move the toy car that will create the largest separation distance.  The magnetic force that will act on the vehicles as he moves the cars away from each other will decrease because magnetic force is inversely proportional to the separating distance between the cars.

Explanation:

Since potential energy increases with increase in the separation distance, Jamal should move the toy car that will create the largest separation distance.The magnetic force that will act on the vehicles as he moves the cars away from each other will decrease because magnetic force is inversely proportional to the separating distance between the cars.

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Answer:

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Explanation:

hope it helps u

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A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What
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Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, \alpha=3.3 rad/s^2

Diameter of the wheel, d=21 cm

Radius of wheel, r=\frac{d}{2}=\frac{21}{2} cm

Radius of wheel, r=\frac{21\times 10^{-2}}{2} m

1m=100 cm

Magnitude of total linear acceleration, a=1.7 m/s^2

We have to find the linear speed  of a  at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration,a_t=\alpha r

a_t=3.3\times \frac{21\times 10^{-2}}{2}

a_t=34.65\times 10^{-2}m/s^2

Radial acceleration,a_r=\frac{v^2}{r}

We know that

a=\sqrt{a^2_t+a^2_r}

Using the formula

1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}

Squaring on both sides

we get

2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}

\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}

v^4=r^2\times 2.7699

v^4=(10.5\times 10^{-2})^2\times 2.7699

v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}

v=0.418 m/s

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

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The separation between the charges was decreased by a factor of 0.2

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