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Eduardwww [97]
3 years ago
14

If a guitar string has a fundamental frequency of 500 hz, which one of the following frequencies can set the string into resonan

t vibration?
Physics
1 answer:
Sav [38]3 years ago
3 0
I had to look for the options and here is my answer:

Given that a guitar string has a basic frequency of 500hz, the frequency that can be set for this into resonant vibration would be 1500 Hz. The resonant vibration here is necessary in musical instruments because it delivers a vibrating system which leads to a higher amplitude at a particular frequency.
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A light spring obeys Hooke's law. The spring's unstretched length is 34.0 cm. One end of the spring is attached to the top of a
sleet_krkn [62]

When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is

∑ F = R - mg = 0

where mg = weight of the mass = (7.00 kg) g = 68.6 N.

It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that

k (0.105 m) = 68.6 N   ⇒   k = (68.6 N) / (0.105 m) ≈ 653 N/m

5 0
2 years ago
An object has a mass of 6kg. calculate it's gpe​
m_a_m_a [10]

Explanation:

When m=<em>mass</em>

G=<em>a</em><em>c</em><em>c</em><em>e</em><em>l</em><em>e</em><em>r</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>d</em><em>u</em><em>e</em><em> </em><em>t</em><em>o</em><em> </em><em>gravity</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>H</em><em>=</em><em>h</em><em>e</em><em>i</em><em>g</em><em>h</em><em>t</em>

<em>U</em><em>s</em><em>i</em><em>n</em><em>g</em><em> </em><em>f</em><em>o</em><em>r</em><em>m</em><em>u</em><em>l</em><em>a</em>

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<em>(</em><em>M</em><em>=</em><em>6</em><em>, </em><em>g</em><em>=</em><em>10</em><em>,</em><em>h</em><em>=</em><em>?</em><em>) </em>

6×10×h

=60joules

7 0
3 years ago
Nvm i got the answer but <br> free points
faust18 [17]

Answer:

thank you 谢谢

Explanation:

8 0
3 years ago
Read 2 more answers
Se coloca una piedra de 600 g en una Honda de 50 cm y se la hace girar a una velocidad de 4 m/s. Dibuja la fuerza que ejerce la
Dima020 [189]

Answer:

Fc = 19.2 N

Explanation:

In this case, the force of the Honda over the rock, is a centripetal force. Then, you have:

F_c=m\frac{v^2}{r}

m: mass of the rock = 600g = 0.6 kg

v: tangential velocity of the Honda = 4m/s

r: radius of the Honda = 50cm = 0.5m

You replace the values of m, r and v in the equation for Fc:

F_c=(0.6kg)\frac{(4m/s)^2}{0.5m}\\\\F_c=19.2N

hence, the force has a magnitude of 19.2 N

If the rock would have more mass the centripetal force would be higher

4 0
3 years ago
Suppose that on earth you can throw a ball vertically upward a distance of 1.20 m. Given that the acceleration of gravity on the
tatuchka [14]

Answer:

7.04 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0

s = Displacement on Earth = 1.2 m

a = Acceleration due to gravity on Moon = 1.67 m/s²

a = Acceleration due to gravity Earth= 9.81 m/s²

Accelration going up is considered as negetive

Initial Velocity of the ball

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 1.2-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 1.2}\\\Rightarrow u=4.85\ m/s

Assuming that the ball is thrown with the same velocity on the Moon, displacement of the ball is

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-4.85^2}{2\times -1.67}\\\Rightarrow s=7.04\ m

The displacement of the ball on the moon is 7.04 m

6 0
4 years ago
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