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Ipatiy [6.2K]
3 years ago
15

A long string is wrapped around a 6.6-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is

then pulled with a constant acceleration of 1.5 m/s2 until 1.3 m of string has been unwound. If the string unwinds without slipping, what is the cylinder's angular speed, in rpm, at this time?
Physics
1 answer:
lys-0071 [83]3 years ago
5 0

Answer:

\omega_f=571.42\ rpm

Explanation:

It is given that,

Diameter of cylinder, d = 6.6 cm

Radius of cylinder, r = 3.3 cm = 0.033 m

Acceleration of the string, a=1.5\ m/s^2

Displacement, d = 1.3 m

The angular acceleration is given by :

\alpha =\dfrac{a}{r}

\alpha =\dfrac{1.5}{0.033}

\alpha =45.46\ rad/s^2

The angular displacement is given by :

\theta=\dfrac{d}{r}

\theta=\dfrac{1.3}{0.033}

\theta=39.39\ rad

Using the third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta

Here, \omega_i=0

\omega_f=\sqrt{2\alpha \theta}

\omega_f=\sqrt{2\times 45.46\times 39.39}

\omega_f=59.84\ rad/s

Since, 1 rad/s = 9.54 rpm

So,

\omega_f=571.42\ rpm

So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.

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Determine the frequency of a sound wave if it has a speed of 350 m/s and a wavelength of 3.80 m.
Eva8 [605]
Since we have , v=f×lambda (wavelength). Where v equals 350m/s and wavelength equals 3.80. so it will become f = v/lambda=350/3.80=92.1052Hz
7 0
3 years ago
An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99
sashaice [31]

<u>Answer:</u> The average atomic mass of the given element is 20.169 amu.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of the isotopes each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i     .....(1)

We are given:

  • For isotope 1:

Mass of isotope 1 = 19.99 amu

Percentage abundance of isotope 1 = 90.92 %

Fractional abundance of isotope 1 = 0.9092

  • For isotope 2:

Mass of isotope 2 = 20.99 amu

Percentage abundance of isotope 2 = 0.26%

Fractional abundance of isotope 2 = 0.0026

  • For isotope 3:

Mass of isotope 3 = 21.99 amu

Percentage abundance of isotope 3 = 8.82%

Fractional abundance of isotope 3 = 0.0882  

Putting values in equation 1, we get:

\text{Average atomic mass}=[(19.99\times 0.9092)+(20.99\times 0.0026)+(21.99\times 0.0882)]

\text{Average atomic mass}=20.169amu

Hence, the average atomic mass of the given element is 20.169 amu.

4 0
3 years ago
A weightlifter lifts a set of weights a vertical distance of 2.00m.If a constant net force of 350 N is exerted on the weights,wh
MrMuchimi

Answer:

<em>W=700 Joule</em>

Explanation:

<u>Physics Work </u>

Is the dot product of the force vector by the displacement vector

W=\vec F \cdot \vec r

When both the force and the displacements are pointed in the same direction, the formula reduces to its scalar version

W=F.d

The weightlifter is applying a net force of 350 N to lift the weights a distance of 2 m, thus the net work done is

W=350\ N\ .\ 2\ m=700\ Joule

4 0
3 years ago
A grandfather clock depends on the period of a pendulum to keep correct time.(ii) Suppose a grandfather clock is calibrated corr
ANEK [815]

The grandfather clock will now run slow (Option A).

<h3>What is Time Period of an oscillation?</h3>
  • The time period of an oscillation refers to the time taken by an object to complete one oscillation.
  • It is the inverse of frequency of oscillation; denoted by "T".

Now,

  • \sqrt{\frac{L}{g}}, where L is the length and g is the gravitational constant, is the formula for a pendulum's period.
  • The period will increase as one climbs a very tall mountain because g will slightly decrease.
  • Due to this and the previous issue, the clock runs slowly and it seems that one second is longer than it actually is.

Hence, the grandfather clock will now run slow (Option A).

To learn more about the time period of an oscillation, refer to the link: brainly.com/question/26449711

#SPJ4

6 0
1 year ago
A steel piano string of mass per unit lenght 4×10-⁴kgm-¹ was struck, if the speed of the sound on the string is 300ms-¹. Calcula
attashe74 [19]

Answer:

36 N

Explanation:

Velocity of a standing wave in a stretched string is:

v = √(T/ρ),

where T is the tension and ρ is the mass per unit length.

300 m/s = √(T / 4×10⁻⁴ kg/m)

T = 36 N

3 0
2 years ago
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