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3 years ago

WHAT IS THE NET FORCE (CENTRIPETAL FORCE) FOR A 1500 KG

1 answer:

5 0

A :-) Given - m = 1500 kg
v = 5 m/s
r = 10 m
Solution -
F = mv^2 by r
F = 1500 x 5 x 5 by 10
( cut 5 and 10 because 5 x 2 = 10 )
F = 1500 x 5 by 2
( cut 2 and 1500 because
2 x 750 = 1500 )
F = 750 x 5
F = 3750 N
.:. The centripetal force is 3750 N.

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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

velocity of wave on the string with greater tension;

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

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

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

Fundamental frequency of wave on the string with greater tension;

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

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