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allsm [11]
1 year ago
7

If a piece of ribbon were tied to a stretched string carrying a transverse wave, then how is the ribbon observed to oscillate?a.

)perpendicular to wave directionb.)both perpendicular to and parallel to wave direction
Physics
1 answer:
noname [10]1 year ago
5 0

The ribbon was observed to oscillate perpendicular to the wave direction. <u>Option A.</u>

<u />

In transverse waves, particles move perpendicular to the direction of wave propagation. Examples of transverse waves are string vibrations and water waves. By moving the Slinky vertically up and down, you can create horizontal transverse waves. Her one type of mechanical wave is the transverse wave.

where the motion of the medium is perpendicular to the direction of energy propagation. A transverse wave is a wave whose vibration is perpendicular to the direction of wave propagation. This is in contrast to longitudinal waves, which propagate in the direction of oscillation. Water waves are an example of transverse waves.

Learn more about The ribbon here:-brainly.com/question/11685530

#SPJ9

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Two equal charges with magnitude Q and Q experience a force of 12.3442 when held at a distance r. What is the force between two
Pepsi [2]

Answer:197.504 N

Explanation:

Given

Two Charges with magnitude Q experience a  force of 12.344 N

at distance r

and we know Electrostatic force is given

F=\frac{kq_1q_2}{r^2}

F=\frac{kQ\cdot Q}{r^2}

F=\frac{kQ^2}{r^2}

Now the magnitude of charge is 2Q and is at a distance of \frac{r}{2}

F'=\frac{k2Q\cdot 2Q}{\frac{r^2}{2^2}}

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4 0
3 years ago
2. A student places an object with a mass of m on a disk at a position r from the center of the disk. The student starts rotatin
ycow [4]

Answer:

\frac{0.065}{r}

Explanation:

The maximum velocity of an object moving in a curve beyond which it will slide off the curve is given by the relationship in equation (1);

v=\sqrt{\mu gr}....................(1)

where \mu is the coefficient of friction between the object and the surface of the curve, g is acceleration due to gravity and r is the radius of the curve.

Given;

v = 0.8m/s

g = 9.81m/s^2

r = ?

\mu=?

In order to solve for \mu, we can simply make it the subject of formula from equation (1) as follows;

v^2=\mu gr\\hence\\\mu=\frac{v^2}{gr}.................(2)

since we were not given the value of r, we can just substitute other known values, then solve and leave the answer in terms of r.

Therefore;

\mu=\frac{0.8^2}{9.81r}

\mu=\frac{0.64}{9.81r}\\\\\mu=\frac{0.065}{r}

8 0
3 years ago
What is the correct explanation of Newton's Third Law of Motion?
viva [34]

Answer:

a

Explanation:

it explains the most, and it is the correct theorem

8 0
2 years ago
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erica [24]

Answer:

gravitational force is a fundamental force and also , it does have a small range and is always an attractive force.

Explanation:

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