These show how something works that can't be seen

Guitar string has an overall length of 1.22 m and a total mass of 3.5 g before being strung on a guitar. Once it is used on the

ikadub [295]

Answer:

Fundamental frequency= 174.5 hz

Explanation:

We know

fundamental frequency= $\frac{velocity}{2 *length}$

velocity = $\sqrt{\frac{tension}{mass per unit length} }$

mass per unit length= $\frac{3.5}{1000*1.22}$ =0.00427 $\frac{kg}{m}$

Now calculating velocity v= $\sqrt{\frac{255}{0.00427} }$

=244.3 $\frac{m}{sec}$

Distance between two nodes is 0.7 m.

Plugging these values into to calculate frequency

f = $\frac{244.3}{2 *0.7}$ =174.5 hz

6 0

3 years ago

Which of the following is an example of pure research? Creating new elements to study their properties Producing plastics that a

koban [17]

Creating new elements in order to study their properties is definitely pure research.
It has no earthly [known] application, since the elements don't even exist, whereas
for each of the other choices, you've stated the desired application as the reason
for the search.

6 0

2 years ago

Which cell part are red blood cells missing

12345 [234]

They lack a cell nucleus.

3 0

3 years ago

How often do the earth's magnetic poles switch?

RSB [31]

Reversals are the rule, not the exception. Earth has settled in the last 20 million years into a pattern of a pole<span> reversal about every 200,000 to 300,000 years, although it has been more than twice that </span>long<span> since the last reversal.</span>

3 0

2 years ago

A 4.0-kg object is supported by an aluminum wire of length 2.0 m and diameter 2.0 mm. How much will the wire stretch?

forsale [732]

Answer:

The extension of the wire is 0.362 mm.

Explanation:

Given;

mass of the object, m = 4.0 kg

length of the aluminum wire, L = 2.0 m

diameter of the wire, d = 2.0 mm

radius of the wire, r = d/2 = 1.0 mm = 0.001 m

The area of the wire is given by;

A = πr²

A = π(0.001)² = 3.142 x 10⁻⁶ m²

The downward force of the object on the wire is given by;

F = mg

F = 4 x 9.8 = 39.2 N

The Young's modulus of aluminum is given by;

$Y = \frac{stress}{strain}\\\\Y = \frac{F/A}{e/L}\\\\Y = \frac{FL}{Ae} \\\\e = \frac{FL}{AY}$

Where;

Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

$e = \frac{FL}{AY} \\\\e = \frac{(39.2)(2)}{(3.142*10^{-6})(69*10^9)} \\\\e = 0.000362 \ m\\\\e = 0.362 \ mm$

Therefore, the extension of the wire is 0.362 mm.

8 0

2 years ago