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Bond [772]
3 years ago
14

Starting from a state of no rotation, a cylinder spins so that any point on its edge has a contant tangential acceleration of 3.

1 m/s2. You keep track of the fractional number of turns N1 made made after 2.9 s and then that N2 made after 10 s. (Those numbers do not have to be integers.) What should be the approximate ratio N2/N1 ?
Physics
1 answer:
Leni [432]3 years ago
3 0

Answer:

Explanation:

tangential acceleration at = 3.1 m  / s²

angular acceleration = tangential accn / radius

= 3.1 / r , r is radius of the cylinder .

IF N₁ be no of rotation in time t

θ = 1/2 α t² , α is angular acceleration , θ is angle in radian covered in time t

2π N₁ = 1/2 (3.1 / r ) x 2.9²

N₁ = 2.0757 / r

similarly we can calculate

2πN₂ = 1/2 (3.1 / r ) x 10²

N₂ = 24.68 / r

N₂ / N₁ = 11.89

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Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init
Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ=(m+\frac{1}{2})\frac{lamda}{n}  ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= \frac{3*lamda}{2}

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= \sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }

δ= \sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}

δ= \sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }

δ= \sqrt{4900+43^2}-\sqrt{4900+23^2}

δ= \sqrt{4900+1849}-\sqrt{4900+529}

δ= \sqrt{6749}-\sqrt{5429}

δ=  82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= \frac{3*lamda}{2}

δ= 8.47

8.47 = \frac{3*lamda}{2}

λ = \frac{8.47*2}{3}

λ = 5.65m

3 0
3 years ago
A trombone plays a C3 note. If the speed of sound in air is 343 m/s and the wavelength of this note is
Umnica [9.8K]

The frequency of note C3 is 131 s^{-1}.

<u>Explanation:</u>

Frequency is the measure of repetition of same thing a certain number of times. So frequency is inversely proportional to the wavelength. As wavelength is distance between two successive crests or troughs in a sound wave.

And frequency is the completion of number of cycles in a given time in sound waves. The frequency and wavelength are inversely proportional to each other with velocity of sound being the proportionality constant.

Thus, here the speed of sound is given as 343 m/s, the wavelength of the note is also given as 2.62 m, then frequency will be as follows:

Frequency=\frac{speed of sound}{Wavelength of note}

Thus,

Frequency = \frac{343 m/s}{2.62 m} = 131 s^{-1}

So the frequency of note C3 is 131 s^{-1}.

3 0
2 years ago
What do you need to know to establish motion
avanturin [10]

Answer:

An object is in motion if it changes position relative to a reference point.

Explanation:

7 0
2 years ago
On both the spring and fall equinoxes, the sun's rays are vertically overhead at the _______
Anni [7]

It is overhead at the equator, it is because the sun ray’s will be moving vertically as this will be directed at the equator. It is because if it moves vertically, it will hit or overhead the equator and this usually happens in spring and fall.

5 0
3 years ago
A merry-go-round is spinning at a rate of 4.04.0 revolutions per minute. Cora is sitting 0.50.5 m from the center of the merry-g
dsp73

Answer:

angular speed of both the children will be same

Explanation:

Rate of revolution of the merry go round is given as

f = 4.04 rev/min

so here we have

f = \frac{4.04}{60} =0.067 rev/s

here we know that angular frequency is given as

\omega = 2\pi f

\omega = 2\pi(0.067)

\omega = 0.42 rad/s

now this is the angular speed of the disc and this speed will remain same for all points lying on the disc

Angular speed do not depends on the distance from the center but it will be same for all positions of the disc

7 0
3 years ago
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