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4 years ago

Suppose you increase your walking speed from 5m/s to 15m/s in a period of 1 s. What is your acceleration?

1 answer:

7 0

Acceleration=change in velocity divided by change in time or $a=\frac{\Delta v}{\Delta t}$

we are given that it goes from 5m/s to 15m/s
$\Delta v=v_f-v_0$
in this case, $\Delta v=15\frac{m}{s}-5\frac{m}{s}=10\frac{m}{s}$

and the change in time or $\Delta t$ is 1 second because this change occurs over a period of 1 second

so the acceleration is $a=\frac{10\frac{m}{s}}{1s}=10\frac{m}{s^2}$
the acceleration is $10\frac{m}{s^2}$

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The period of the wave is 4.35 ms. The sound waves are called longitudinal waves

Explanation:

The period of a wave is related to its frequency by the equation:

$T=\frac{1}{f}$

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For the bee in this problem, the frequency of the sound wave emitted by it is

$f=2.3 \cdot 10^2 Hz = 230 Hz$

Therefore, the period of the sound wave is

$T=\frac{1}{230}=4.35\cdot 10^{-3}s = 4.35 ms$

The sound wave is a type of wave called longitudinal wave. In longitudinal waves, the oscillation of the medium occurs in a direction parallel to the direction of motion of the wave: therefore in a sound wave, the particle of the medium (air, in this case) oscillate back and forth along the direction of propagation of the wave, forming alternating areas of higher density of particles (called compressions) and of lower density of particle (called rarefactions).

The other type of wave, instead, is called transverse wave. In a transverse wave, the oscillation of the wave occurs in a direction perpendicular to the direction of motion of the wave. An example of transverse waves are the electromagnetic waves, which consists of electric field and magnetic fields that vibrate in a plane perpendicular to the direction of motion of the wave itself.

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brainly.com/question/5354733

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3 years ago

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3 years ago

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Gala2k [10]

Answer:

When viewed from above, the current in the coil should point towards the top-right corner of the picture.

Explanation:

The current in this coil have only two possible directions: clockwise or counter-clockwise. However, since the diagram shows the coil from above, not from a cross-section, just saying clockwise or counter-clockwise might be ambiguous. The statement that the current is directed towards the top-right corner of the picture is equivalent to saying that when viewed from the lower-right corner of this diagram, the current in the coil is moving clockwise.

Note that at the center of this picture, the current is parallel to the magnetic field- there will be no force on the coil at that position. On the other hand, (also when viewed from above,) at the top-right corner and the lower-left corner of the coil, the current in the coil will be perpendicular to the magnetic field. That's where the force on the coil will be the strongest.

With that in mind, apply the right-hand rule to find the direction of the force on the coil in each of the two possibilities.

Assume that when viewed from above, the current is flowing towards the top-right corner of the picture. Consider the wire near the top-right corner of this coil (as viewed above on this picture.) The current will be going into the picture into the magnetic field. By the right-hand rule, the current on the wire near that point should be pointing towards the bottom of this picture. (Point fingers on the right hand in the direction of the current $I$ . Rotate the right hand such that when curling the fingers, they point in the direction of the magnetic field $B$ . The direction of the right thumb should now point in the direction of the force on the wire $F$ .)

Based on the same assumption, the current in the wires near the bottom left corner of this coil will be pointing out of the picture. By the right hand rule, the magnetic force on the coil in that region should be pointing towards the top of this picture. Combing these two forces, the coil would indeed be rotating around the center of this picture in the direction shown in the diagram.

It can also be shown that if the current points towards the bottom left corner of the picture when viewed from above, the coil will be rotating about the center of this picture in the opposite direction.

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3 years ago

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Answer:

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