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lisabon 2012 [21]
3 years ago
5

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

guitar, there is a distance of 70 cm between fixed end points. The guitar string is tightened to a tension of 255 N.
What is the frequency of the fundamental wave on the guitar string?
Physics
1 answer:
ikadub [295]3 years ago
6 0

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

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in a certain experiment, a radio transmitter emits sinusoidal electromagnetic waves of frequency 105.0 mhz in opposite direction
Romashka [77]

As a result, the hollow is 10.90 meters long and the distance between the nodal planes is 1.36 meters.

<h3>Explain electromagnetic waves.</h3>

The oscillations between an electric field and a magnetic field produce waves known as electromagnetic waves, or EM waves.

By definition, we understand that the frequency equals,

f = c/λ

where,

λ = wavelength

c= Speed of light

λ = 2L / n

While the wavelength is equal to,

Where,

L = Length

n = Number of antinodes/nodes

PART A) We know that the first component's wavelength is 110 MHz, so

λ = c/ f

λ = 3*10^8 / 11*10^6

λ = 1.36m

Therefore the distance between the nodal planes is 1.36m

PART B) For this part we need to find the Length through the number of nodes (8) and the wavelength, that is,

λ` = 2l /n

L = 8*2.72/ 2

L = 10.90m

Therefore the length of the cavity is 10.90m.

To know more about electromagnetic waves visit:-

brainly.com/question/3101711

#SPJ4

7 0
1 year ago
because lighting heats up the surrounding air slowly, it realationship to thunder is currently unknown true or false​
NeTakaya

Answer:

False

Explanation:

8 0
3 years ago
A pitcher releases a fastball that moves toward home plate.
Vsevolod [243]

The two additional forces that act on the ball as it travels between the pitcher and the home plate are air resistance and gravity.

<h3>What are the forces that affect object in motion;</h3>
  • Air resistance: this is the force that oppose the motion of an object in air due to frictional force
  • Gravity: this is the force due to weight of the object and acts downwards.

The two additional forces that act on the ball as it travels between the pitcher and the home plate include:

  1. Air resistance and
  2. Gravitational force  

<h3>How the forces affect the motion of the ball</h3>
  • Air resistance oppose the motion of the ball as it travels in air.
  • Gravity is the force due to weight of the ball and acts downwards.

Learn more about forces on object in motion here: brainly.com/question/10454047

5 0
2 years ago
If a bullet loses 1/nth of its velocity while passing through a plank,then no of planks required to stop the bullet is?
Novay_Z [31]
Based on the answer provided, it seems the writer wanted you to assume that the energy loss per plank is constant. This is not the same as the bullet losing <span><span>1/nth</span><span>1/nth</span></span><span> of its velocity per plank (however, the fact that the question does not mention this assumption arguably makes the question ambiguous).

</span><span>With this assumption, the energy loss becomes
</span><span>
ΔE = <span>1/2 </span>m<span>v2 </span>− <span>1/2 </span>m <span><span>(<span>v−<span>v/n</span></span>) </span><span>2
</span></span></span>
and the number of planks <span>NN</span><span> becomes
</span>
N = <span><span><span>1/2</span>m<span>v2 /</span></span><span>ΔE </span></span>= <span><span>n2/ </span><span>2n−1
</span></span>
Otherwise, if you assume that the bullet loses <span><span>1/<span>nth</span></span><span>1/<span>nth</span></span></span><span> of its velocity per plank, then the answer is </span><span><span>N=∞</span></span><span><span>

</span>
</span>
8 0
3 years ago
Two identical metal spheres a and b are connected by a plastic rod. both are initially neutral. 1.0×1012 electrons are added to
Ann [662]
The plastic rod is made of insulator (plastic), so it does not allow charges moving from one sphere to another. This means that all the electrons given to sphere A will remain on sphere A.

The number of electrons initially given to sphere A is N=1.0 \cdot 10^{12}, and since the charge of 1 electron is e=-1.6 \cdot 10^{-19} C, the net charge left on sphere A after the removal of the rod will be
Q=N e = 1.0 \cdot 10^{12}(-1.6 \cdot 10^{-19} C)=-1.6.0 \cdot 10^{-7} C
8 0
3 years ago
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