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ArbitrLikvidat [17]

3 years ago

14

0.25-kg block oscillates on the end of a spring with a spring constant of 200 N/m. If the oscillations is started by elongating

the spring 0.15 m and giving the block a speed of 3.0 m/s, then the maximum speed of the block is A :

1 answer:

ValentinkaMS [17]3 years ago

5 0

Answer:

5.2m/s

Explanation:

Plss see attached file

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<u>Explanation:</u>

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$\theta=8^{\circ}$

Magnetic field of earth, $B_{\text {earth}}=2 \times 10^{-5}\ \text {tesla}$

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The compass needle moves toward a direction of magnetic field. The current in wire makes a magnetic field in available space where the compass is on the ground. The vector sum of the Earth's magnetic field and the wire's magnetic field represents the net magnetic field, as shown in the attached drawing, expressing the angle:

$\tan \theta=\frac{B_{\text {wire}}}{B_{\text {earth}}}$

By substituting the given values, we get

$B_{\text {wire}}=\tan \theta \times B_{\text {earth}}=\tan 8^{\circ} \times 2 \times 10^{-5}=0.1405 \times 2 \times 10^{-5}$

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