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Ainat [17]
3 years ago
8

The amplitude of a particular wave is 4.0 m. the top-to-bottom distance of the disturbance is

Physics
1 answer:
fenix001 [56]3 years ago
5 0

The amplitude of a wave is the distance from the center of the wave to one of its peaks or one of its troughs. The distance from center to peak or center to trough is the same, therefore the top-to-bottom distance is simply twice the wave's amplitude, 8.0m

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erma4kov [3.2K]
A transformer is an electrical device that transfers energy between two or more circuits through electromagnetic induction. Transformers are mainly used to increase or decrease voltage in power lines or electric power applications. So your answer would be D. transformer
4 0
3 years ago
Read 2 more answers
A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so sma
mel-nik [20]

The angular speed of the merry-go-round reduced more as the sandbag is

placed further from the axis than increasing the mass of the sandbag.

The rank from largest to smallest angular speed is presented as follows;

[m = 10 kg, r = 0.25·R]

              {} ⇩

[m = 20 kg, r = 0.25·R]

              {} ⇩

[m = 10 kg, r = 0.5·R]

              {} ⇩

[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]

              {} ⇩

[m = 10 kg, r = 1.0·R]

Reasons:

The given combination in the question as obtained from a similar question online are;

<em>1: m = 20 kg, r = 0.25·R</em>

<em>2: m = 10 kg, r = 1.0·R</em>

<em>3: m = 10 kg, r = 0.25·R</em>

<em>4: m = 15 kg, r = 0.75·R</em>

<em>5: m = 10 kg, r = 0.5·R</em>

<em>6: m = 40 kg, r = 0.25·R</em>

According to the principle of conservation of angular momentum, we have;

I_i \cdot \omega _i = I_f \cdot \omega _f

The moment of inertia of the merry-go-round, I_m = 0.5·M·R²

Moment of inertia of the sandbag = m·r²

Therefore;

0.5·M·R²·\omega _i = (0.5·M·R² + m·r²)·\omega _f

Given that 0.5·M·R²·\omega _i is constant, as the value of  m·r² increases, the value of \omega _f decreases.

The values of m·r² for each combination are;

Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²

Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²

Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²

Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²

Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²

Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²

Therefore, the rank from largest to smallest angular speed is as follows;

Combination 3 > Combination 1 > Combination 5 = Combination 6 >

Combination 2

Which gives;

[<u>m = 10 kg, r = 0.25·R</u>] > [<u>m = 20 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 0.5·R</u>] > [<u>m = </u>

<u>10 kg, r = 0.5·R</u>] = [<u>m = 40 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 1.0·R</u>].

Learn more here:

brainly.com/question/15188750

6 0
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A 50-kg box is being pushed along a horizontal surface. The coefficient of static friction between the box and the ground is 0.6
dsp73

To solve the exercise it is necessary to apply the concepts related to Newton's Second Law, as well as the definition of Weight and Friction Force.

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I think it is 4 cm because if you look hard enough and work the problem the number pops in you head. Lol
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