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Natasha2012 [34]
3 years ago
11

The guitarist shortens the oscillating length of the properly tuned D-string by 0.15 m by pressing on the string with a finger.

What is the fundamental frequency, in hertz, of the new tone that is produced when the string is plucked?
Physics
1 answer:
Greeley [361]3 years ago
3 0

Answer:

The fundamental frequency is "190.52 Hz".

Explanation:

The given question is incomplete, please find attachment of the complete problem.

The given values are:

Frequency,

f = 146.8 Hz

Length of D-string,

L = 0.61

As we know,

⇒ f = \frac{v}{2L}

On putting the given values, we get

⇒ 146.8 = \frac{v}{(2\times 0.61)}

⇒       v=146.8\times 1.22

⇒       v=179 \ m/s

So,

The new length will be:

⇒  f = \frac{179}{(2\times 0.47)}

⇒     =\frac{179}{0.94}

⇒     =190.52 \ Hz

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An object starts at 11 m/s with an acceleration of 12 m/s? What is its velocity after 4.5 seconds?
marshall27 [118]

Answer:

65 m/s

Explanation:

v=v0+at <=> v = 11 + 12 t ∧ t = 4.5 s <=> v = 11 + 12×4.5 <=> v = 65 m/s

5 0
3 years ago
For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at ө = 45°. Substant
myrzilka [38]

First let us imagine the projectile launched at initial velocity V and at angle θ relative to the horizontal. (ignore wind resistance)

Vertical component y:

The initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum height of h, the vertical velocity will be 0, therefore the time t taken to attain this maximum height is:

h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g

where g is  acceleration due to gravity

Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g] 
D = (2V²sinθ cosθ)/g 
 D = (V²sin2θ)/g

In order for D (horizontal distance) to be maximum, dD/dθ = 0
That is,

2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2  or  θ = π/4


Therefore it is now<span> shown that the maximum horizontal travelled is attained when the launch angle is π/4 radians, or 45°.</span>

6 0
3 years ago
A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done
Elena-2011 [213]

Answer:

The work done by gravity during the roll is 490.6 J

Explanation:

The work (W) is:

W = F*d

<em>Where</em>:

F: is the force

d: is the displacement = 20 m

The force is equal to the weight (W) in the x component:

F = W_{x} = mgsin(\theta)

<em>Where:</em>

m: is the mass of the bowling ball = 5 kg

g: is the gravity = 9.81 m/s²    

θ: is the degree angle to the horizontal = 30°        

F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N    

Now, we can find the work:

W = F*d = 24.53 N*20 m = 490.6 J      

Therefore, the work done by gravity during the roll is 490.6 J.

I hope it helps you!

6 0
3 years ago
The current through a 10 ohm resistor connected to a 120 volt power supply is
Leno4ka [110]

Answer:I=12 A

Explanation:

Given

Resistance R=10 \Omega

Voltage V=120 V

According to ohm's law current through a conductor is directly proportional to the voltage applied.

V\propto I

V=IR

where V=Voltage

I=Current

R=resistance

I=\frac{V}{R}

I=\frac{120}{10}

I=12 A

4 0
3 years ago
The radioactive 60co isotope is used in nuclear medicine to treat certain types of cancer. Calculate the wavelength and frequenc
Ivanshal [37]

1. Frequency: 3.23\cdot 10^{20} Hz

The energy given is the energy per mole of particles:

E=1.29\cdot 10^{11} J/mol

1 mole contains a number of Avogadro of particles, N_A, equal to

N_A=6.022\cdot 10^{23} particles

So, by setting the following proportion, we can calculate the energy of a single photon:

1.29 \cdot 10^{11} J/mol : 6.022 \cdot 10^{23} ph/mol = E_1 : 1 ph\\E_1 = \frac{(1.29\cdot 10^{11} J/mol)(1 ph)}{6.022\cdot 10^{23} ph/mol}=2.14\cdot 10^{-13} J

This is the energy of a single photon; now we can calculate its frequency by using the formula:

E_1 = hf

where

h=6.63\cdot 10^{-34} Js is the Planck's constant

f is the photon frequency

Solving for f, we find

f=\frac{E_1}{h}=\frac{2.14\cdot 10^{-13} J}{6.63\cdot 10^{-34} Js}=3.23\cdot 10^{20} Hz

2. Wavelength: 9.29\cdot 10^{-13} m

The wavelength of the photon is given by the equation:

\lambda=\frac{c}{f}

where

c=3\cdot 10^8 m/s

is the speed of the photon (the speed of light). Substituting,

\lambda=\frac{3 \cdot 10^8 m/s}{3.23\cdot 10^{20} Hz}=9.29\cdot 10^{-13} m

6 0
3 years ago
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