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VLD [36.1K]
3 years ago
6

5. Sandor fills a bucket with water and whirls it in a vertical circle to demonstrate that the

Physics
1 answer:
just olya [345]3 years ago
3 0

Given that,

radius = 1.24 m

According to question,

The rope cannot push outwards. It must always have some slight tension or the bucket will fall.

We need to calculate the tension in the rope

At the top the force of gravity is

F=mg

The force needed to move the bucket in a circle is centripetal force.

So, if mg is ever greater than centripetal force then the bucket and the contents will start to fall.

The rope have a tension of less than zero.

We need to calculate the velocity of swing bucket

Using centripetal force

F=\dfrac{mv^2}{r}

mg=\dfrac{mv^2}{r}

g=\dfrac{v^2}{r}

v^2=gr

v=\sqrt{gr}

Put the value into the formula

v=\sqrt{9.8\times1.24}

v=3.49\ m/s

Hence, The minimum tension in the rope is less than zero .

The bucket swings with the velocity of 3.49 m/s.

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What is the acceleration of a 10 kg mass pushed by a 5 N force?
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Differentiate scalar & vector quantity?
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\textbf{Hello Friend}


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5 0
3 years ago
Two taut strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times g
ollegr [7]

Answer:

The  beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

Explanation:

Given;

velocity of wave on the string with lower tension, v₁ = 35.2 m/s

the fundamental frequency of the string, F₁ = 258 Hz

<u>velocity of wave on the string with greater tension;</u>

v_1 = \sqrt{\frac{T_1}{\mu }

where;

v₁ is the velocity of wave on the string with lower tension

T₁ is tension on the string

μ is mass per unit length

v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} =  \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}

Where;

T₁ lower tension

T₂ greater tension

v₁ velocity of wave in string with lower tension

v₂ velocity of wave in string with greater tension

From the given question;

T₂ = 1.1 T₁

v_2^2 = \frac{T_2v_1^2}{T_1}  \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s

<u>Fundamental frequency of wave on the string with greater tension;</u>

<u />f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1}  =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz<u />

Beat frequency = F₂ - F₁

                          = 270.6 - 258

                          = 12.6 Hz

Therefore, the  beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

6 0
3 years ago
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