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RSB [31]
2 years ago
6

A particular string resonates in four loops at a frequency of 320 Hz . Name at least three other (smaller) frequencies at which

it will resonate. Express your answers using two significant figures separated by commas.
Physics
1 answer:
goldfiish [28.3K]2 years ago
6 0

Answer:

160 Hz  ,  240 Hz  , 400 Hz

Explanation:

Given that

Frequency of forth harmonic is 320 Hz.

Lets take fundamental frequency = f₁

f_1=\dfrac{320}{4}\ Hz

f₁=80 Hz

Frequency of first harmonic = f₂

f₂=2 f₁

f₂ =2 x 80 = 160 Hz

Frequency of second harmonic = f₃

f₃= 3 f₁=3 x 80 = 240 Hz

Frequency of fifth harmonic = f₅

f₅=  5 f₁= 5 x 80 = 400 Hz

Three frequencies are as follows

160 Hz  ,  240 Hz  , 400 Hz

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You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph from due north.
Afina-wow [57]

Answer:

<em>a) 17.05 mph</em>

<em>b) 54.7°  northeast direction</em>

<em>c) 10.71 mph</em>

<em>The direction is -22.58° relative to the east.</em>

<em></em>

<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>

Explanation:

The question is a little confusing but, I guess the correct question should be;

You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph due north.

a) What is your airspeed?

b) What angle (direction) are you flying?

c) The wind increases to 14 mph from north. Now what is your airspeed and what direction are you flying? If your destination is to the northeast, how would you change your speed or direction so you might make it there?

NB: The difference in the question and my suggestion is highlighted boldly.

Your speed = 14 mph

direction is 45° northeast

Th wind speed = 4 mph

direction is north

We resolve the your speed and the wind speed into the horizontal and vertical components

For vertical the component component

V_{y} = 14(sin 45) + 4 = 9.89 + 4 = 13.89 mph

For the horizontal speed component

V_{x} = 14(cos 45) + 0 = 9.89 + 0 = 9.89 mph

Resultant speed = \sqrt{V^{2} _{y}+V^{2} _{x}  }

==> \sqrt{13.89^{2} +9.89^{2}   } = <em>17.05 mph  This is your airspeed</em>

b) To get your direction, we use

tan ∅ = V_{y} /V_{x}

tan ∅ = 13.89/9.89 = 1.413

∅ = tan^{-1}(1.413) = <em>54.7°  northeast direction</em>

c) If the wind increases to 14 mph from the north, then it means the wind blows due south. As before, only the vertical component is affected .

In this case,

V_{y} = 14(sin 45) - 14 = 9.89 - 14 = -4.11 mph

Resultant speed = \sqrt{V^{2} _{y}+V^{2} _{x}  }

==> \sqrt{4.11^{2} +9.89^{2}   } = <em>10.71 mph  This is your airspeed</em>

Your direction will be,

tan ∅ = V_{y} /V_{x}

tan ∅ = -4.11/9.89 = -0.416

∅ = tan^{-1}(-0.416) =<em> -22.58°  this is the angle you'll travel relative to the east.</em>

<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>

5 0
2 years ago
Maria throws an apple vertically upward from a height of 1.3 m with an initial velocity of +2.8 m/s. Will the apple reach a frie
evablogger [386]

Answer:

No, the apple will reach 4.20041 m below the tree house.

Explanation:

t = Time taken

u = Initial velocity = 2.8 m/s

v = Final velocity = 0

s = Displacement

g = Acceleration due to gravity = -9.81 m/s² = a (negative as it is going up)

Equation of motion

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-2.8^2}{2\times -9.81}\\\Rightarrow s=0.39959\ m

The height to which the apple above the point of release will reach is 0.39959 m

From the ground the distance will be 1.3+0.39959 = 1.69959 m

Distance from the tree house = 5.9-1.69959 = 4.20041 m

No, the apple will reach 4.20041 m below the tree house.

The values in the option do not reflect the answer.

5 0
2 years ago
A small light cylinder and a large heavy cylinder are released at the same time and roll down a ramp without slipping. Which one
forsale [732]

Answer:

C. Both reach the bottom at the same time.

Explanation:

For a rolling object down an inclined plane , the acceleration is given below

a = g sinθ / (1 + k² / r² )

θ is angle of inclination , k is radius of gyration , r is radius of the cylinder

For cylindrical object

k² / r² = 1/2

acceleration =  g sinθ  /( 1 + 1/2 )

= 2 g sinθ / 3  

Since it does not depend upon either mass or radius , acceleration of both the cylinder will be equal . Hence they will reach the bottom simultaneously.

6 0
2 years ago
Arwan finds a piece of quartz while hiking in the mountains. When he returns to school, he takes it to his science teacher. She
bekas [8.4K]

it's 303.4 cm3. i just took the test

8 0
2 years ago
Read 2 more answers
You are called as an expert witness to analyze the following auto accident: Car B, of mass 2000 kg, was stopped at a red light w
OLga [1]

Answer:

a). va=17.23 \frac{m}{s} or 38.54 mph

b). v=38.54 mph and limit is 35 mph

c). Completely inelastic

d). Eka=192.967 kJ

Ekt=76.071 kJ

Explanation:

m_{a}=1300kg\\m_{b}=2000kg\\x_{f}=7.25m\\u_{k}=0.65

The motion is an inelastic collision so

m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}

The force of the motion is contrarest by the force of friction so

F-F_{uk} =0\\F=F_{uk}\\F_{uk}=u_{k}*m*g\\F=m*a\\a=\frac{F}{m}\\ a=\frac{F_{uk}}{m}\\a=\frac{u_{k}*m*g}{m}\\a=u_{k}*g\\a=0.65*9.8\frac{m}{s^{2}} \\a=6.39\frac{m}{s^{2}}

Now with the acceleration can find the time and the velocity final that make the distance 7.25m being united

x_{f}=x_{o}+v_{o}*t+2*a*t^{2}\\x_{o}=0\\v_{o}=0\\x_{f}=2*a*t^{2}\\t^{2}=\frac{x_{f}}{2*a}\\t=\sqrt{\frac{7.25m}{6.37\frac{m}{s^{2} } } } \\t=1.06s

So the velocity final can be find using this time

v_{f}=v_{o}+a*t\\v_{o}=0\\v_{f}=6.37\frac{m}{s^{2} } *1.06s\\v_{f}=6.79 \frac{m}{s}

a).

Replacing in the first equation the final velocity can find the initial velocity

m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}

v_{b}=0

v_{a}= \frac{(m_{a}+m_{b)*v_{f}}}{m_{a}}\\v_{a}= \frac{(1300+2000)*6.37}{1300}\\v_{a}=17.23 \frac{m}{s}

b).

35mph*\frac{1m}{0.000621371mi} *\frac{1h}{3600s}=15.646\frac{m}{s}

Velocity limit in m/s is 15.646 m/s and the initial velocity is 17.23 m/s

so is exceeding the speed limit in about 1.58 m/s

or in miles per hour

3.5 mph

c).

The collision is complete inelastic because any mass can be returned to the original mass, so even they are no the same mass however in the moment they move the distance 7.25m as a same mass the motion is considered completely inelastic

d).

Ek=\frac{1}{2}*m*(v)^{2}\\  Eka=\frac{1}{2}*1300kg*(17.23\frac{m}{s})^{2}\\Eka=192.967 kJ\\Ekt=\frac{1}{2}*m*(v)^{2}\\Ekt=\frac{1}{2}*3300kg*(6.79\frac{m}{s})^{2}\\Ekt=76.071 kJ

8 0
2 years ago
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