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3 years ago

8

Determine the weight of an average physical science textbook whose mass is 3.1 kilograms. The acceleration due to gravity is 9.8

m/s2.

1 answer:

Mrrafil [7]3 years ago

5 0

Given:

mass is 3.1 kilograms

The acceleration due to gravity is 9.8m/s2

Required:

Weight

Solution:

W = mg

W = (3.1 kilograms)( 9.8m/s2)

W = 30.38 Newtons

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Standing waves are created in the four strings shown in Figure 25. All strings have the same mass per unit length and are under

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The rank of the frequencies from largest to smallest is

The largest frequency of oscillation is given by the string in option D

The second largest frequency of oscillation is given by the string in option B

The third largest frequency of oscillation is given by the string in option A

The smallest frequency of oscillation is given by the string in option C

Explanation:

The given parameters are;

The mass per unit length of all string, m/L = Constant

The tension of all the string, T = Constant

The frequency of oscillation, f, of a string is given as follows;

$f = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L}$

Where;

T = The tension in the string

m = The mass of the string

L = The length of the string

n = The number of overtones

$Therefore, \ {\sqrt{\dfrac{T}{m/L} } } = Constant \ for \ all \ strings = K$

For the string in option A, the length, L = 27 cm, n = 3 we have;

$f_A = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(3 + 1) \times K }{2 \times 27} = \dfrac{2 \times K}{27} \approx 0.07407 \cdot K$

For the string in option B, the length, L = 30 cm, n = 4 we have;

$f_B = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(4 + 1) \times K }{2 \times 30} = \dfrac{ K}{12} \approx 0.08 \overline 3\cdot K$

For the string in option C, the length, L = 30 cm, n = 3 we have;

$f_C = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(3 + 1) \times K }{2 \times 30} = \dfrac{K}{15} \approx 0.0 \overline 6 \cdot K$

For the string in option D, the length, L = 24 cm, n = 4 we have;

$f_D = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(4 + 1) \times K }{2 \times 24} = \dfrac{5 \times K}{48} \approx 0.1041 \overline 6 \cdot K$

Therefore, we have the rank of the frequency of oscillations of th strings from largest to smallest given as follows;

1 ) $f_D$ 2) $f_B$ 3) $f_A$ 4) $f_C$

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