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3 years ago

9

How do you find circular motion?

2 answers:

Ksivusya [100]3 years ago

7 0

Answer:

2πr

as we know that the circumference of a circle is 2 * π * r, the arc length of the whole circle is 2πr units. Then using the formula s = r * theta, the angle theta equals the arc length divided by the radius. Therefore theta = 2πr/r = 2π radians.

I hope it will helpful for you

mark as brainest answer

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MatroZZZ [7]3 years ago

6 0

Answer:

Explanation:

By multiplying the rotational frequency with the circumference we can determine the average speed of the object. The circular velocity formula is expressed as, vc = 2 πr / T. Where in, r denotes the radius of the circular orbit. T is time period.

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The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz. Find the possible range of wavelengths in ai

taurus [48]

Answer:

The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

Explanation:

Given that,

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.

The speed of sound in air is 343 m/s.

To find,

The wavelength range for the corresponding frequency.

Solution,

The speed of sound is given by the following relation as :

$v=f_1\lambda_1$

Wavelength for f = 45 Hz is,

$\lambda_1=\dfrac{v}{f_1}$

$\lambda_1=\dfrac{343}{45}=7.62\ m$

Wavelength for f = 375 Hz is,

$\lambda_2=\dfrac{v}{f_2}$

$\lambda_2=\dfrac{343}{375}=0.914\ m/s$

So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

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3 years ago

His eyes are 1.64 m above the floor and the top of his head is 0.14 m higher. Find the height (in m) above the floor of the top

Aloiza [94]

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A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is

Lana71 [14]

Answer:

a

$m = 0.169 \ kg$

b

$|v_{max} |= 0.5653 \ m/s$

Explanation:

From the question we are told that

The spring constant is $k = 14 \ N/m$

The maximum extension of the spring is $A = 6.0 \ cm = 0.06 \ m$

The number of oscillation is $n = 30$

The time taken is $t = 20 \ s$

Generally the the angular speed of this oscillations is mathematically represented as

$w = \frac{2 \pi}{T}$

where T is the period which is mathematically represented as

$T = \frac{t}{n}$

substituting values

$T = \frac{20}{30 }$

$T = 0.667 \ s$

Thus

$w = \frac{2 * 3.142 }{ 0.667}$

$w = 9.421 \ rad/s$

this angular speed can also be represented mathematically as

$w = \sqrt{\frac{k}{m} }$

=> $m =\frac{k }{w^2}$

substituting values

$m =\frac{ 15 }{(9.421)^2}$

$m = 0.169 \ kg$

In SHM (simple harmonic motion )the equation for velocity is mathematically represented as

$v = - Awsin (wt)$

The velocity is maximum when $wt = \(90^o) \ or \ 1.5708\ rad$

$v_{max} = - A* w$

=> $|v_{max} |= A* w$

=> $|v_{max} |= 0.06 * 9.421$

=> $|v_{max} |= 0.5653 \ m/s$

5 0

3 years ago