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german
3 years ago
9

When a guitar string is plucked, the guitar string oscillates as a result of waves moving through the string. The vibrations of

the string cause the air molecules to oscillate, forming sound waves. The frequency of the sound waves is equal to the frequency of the vibrating string. Is the wavelength of the sound wave always equal to the wavelength of the waves on the string
Physics
1 answer:
oksian1 [2.3K]3 years ago
4 0

Answer:

Explanation:

Wave produced in string and waves produced in air are different . The only similarity is that their frequency are equal . Otherwise , no similarity . One is transverse ( on wire ) , the other is longitudinal ( air ) . Their velocities too are different .

velocity = wavelength x frequency

if frequency is constant

wavelength  ∝  velocity

wavelength is proportional to velocity . Since their velocities are different , their wavelength too will be different.

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When a pitcher throws a softball to a catcher, the vibration of the atoms that make up the softball is ____________ energy, whil
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The vibration is thermal energy ("heat" energy which every object possesses).
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Grasses, shrubs and trees are called producers because they make ______a. water_____b. carbon dioxide_____c. minerals_____d. foo
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B?
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3 years ago
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The product nuclei may or may not be _______
Vsevolod [243]

Answer:

i have no clue

Explanation:

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3 years ago
Three heavy rods are all made of the same uniform material. These rods have lengths 3 m, 4 m, and 5 m, and compose the sides of
Sonbull [250]

Answer:

[ 2.67 , 1 ] m

Explanation:

Given:-

- The side lengths of the rods are as follows:

                             a = 4 m , b = 4 m , c = 5 m

                             a = Base , b = Perpendicular , c = Hypotenuse

- All rods are made of same material with uniform density. With  

Find:-

Find the coordinates of the center of mass of the triangle.

Solution:-

- The center of mass of any triangle is at the intersection of its medians.

- So let’s say we have a triangle with vertices at points (0,0) , (a,0) , and (0,b).

  • Median from (0,0) to midpoint (a/2,b/2) of opposite side has equation:

                                       bx−ay=0

  • Median from (a,0) to midpoint (0,b/2) of opposite side has equation:

                                      bx+2ay=ab

  • Median from (0,b) to midpoint (a/2,0) of opposite side has equation:

                                     2bx+ay=ab

  • Solve all three equations simultaneously:

                                     bx−ay=0  , bx = ay

                                     ay + 2ay = ab , 3ay = ab , y = b/3

                                     bx = b/3

                                     x = a / 3

  • So the distance from the median to each leg of the triangle is 1/3 length of other leg.

- So the coordinates of the centroid for right angle triangle would be:

                                   [ 2a/3 , b/3 ]

                                   [ 2.67 , 1 ] m

                                                         

                                 

3 0
3 years ago
A cylinder of mass 14.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed o
aev [14]

Answer:

a) 567J

b) 283.5J

c)850.5J

Explanation:

The expression for the translational kinetic energy is,

E_r = \frac{1}{2} mv^2

Substitute,

14kg for m

9m/s for v

E_r = \frac{1}{2} (14) (9)^2\\= 567J

The translational kinetic energy of the center of mass is 567J

(B)

The expression for the rotational kinetic energy is,

E_R = \frac{1}{2} Iw^2

The expression for the moment of inertia of the cylinder is,

I = \frac{1}{2} mr^2

The expression for angular velocity is,

w = \frac{v}{r}

substitute

1/2mr² for I

and vr for w

in equation for rotational kinetic energy as follows:

E_R = (\frac{1}{2}) (\frac{1}{2} mr^2)(\frac{v}{r} )^2

= \frac{mv^2}{4}

E_R = \frac{14 \times 9^2 }{4} \\\\= 283.5J

The rotational kinetic energy of the center of mass is 283.5J

(c)

The expression for the total energy is,

E = E_r + E_R\\\\

substitute 567J for E(r) and 283.5J for E(R)

E = 567J + 283.5\\= 850.5J

The total energy of the cylinder is 850.5J

6 0
3 years ago
Read 2 more answers
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