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UNO [17]
3 years ago
12

3. Una cuerda de guitarra tiene 60 cm de longitud y una masa de 0.05 kg de masa. Si se tensiona mediante una fuerza de 20 N. La

frecuencia del primer armónico o tono fundamental es
Physics
1 answer:
jok3333 [9.3K]3 years ago
4 0

Answer:

f1 = 12.90 Hz

Explanation:

To calculate the first harmonic frequency you use the following formula for n = 1:

f_n=\frac{n}{2L}\sqrt{\frac{T}{M/L}}

f_1=\frac{1}{2L}\sqrt{\frac{T}{M/L}}    ( 1 )

It is necessary that the unist are in meters, then you have:

L: length of the string = 60cm = 0.6m

M: mass of the string = 0.05kg

T: tension on the string = 20 N

you replace the values of L, M and T in the expression (1) for getting f1:

f_1=\frac{1}{2(0.6m)}\sqrt{\frac{20N}{0.05kg/0.6m}}=12.90\ Hz

Hence, the first harmonic has a frequency of 12.90 Hz

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Yall have helped a lot i just need help on this then ill be done for a while
Lubov Fominskaja [6]

Answer:

f(x)=a(x - h)2 + k

Much like a linear function, k works like b in the slope-intercept formula. Like where add or subtract b would determine where the line crosses, in the linear, k determines the vertex of the parabola. If you're going to go up 2, then you need to add 2.

The h determines the movement horizontally. what you put in h determines if it moves left or right. To adjust this, you need to find the number to make the parentheses equal 0 when x equals -2 (because moving the vertex point to the left means subtraction/negatives):

x - h = 0

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h = -2

So the function ends up looking like:

f(x)=a(x - (-2))2 + 2

Subtracting a negative cancels the signs out to make a positive:

f(x)=a(x + 2)2 + 2Explanation:

6 0
3 years ago
A solid brass cylinder and a solid wood cylinder have the same radius and mass (the wood cylinder is longer). Released together
Lena [83]

Answer:

a. They will be tie

b. Win the wood cylinder

Explanation:

a.

The both cylinders will reach the bottom at the same time notice the relation in the equation in indepent of the length and both have the same radius and the same rotational inertia.

I=\frac{1}{2}*m*r^2

a=\frac{g*sin(\beta)}{1+I_{com}/m*r^2}

So both will be tie

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a_{brass}=\frac{g*sin(\beta)}{1+I_{brass}/m*r^2}=a_{wood}=\frac{g*sin(\beta)}{1+I_{wood}/m*r^2}

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a_{brass}

So the wood win the race

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