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adoni [48]
3 years ago
6

A resonance tube can be used to measured the speed of sound in air. A tuning fork is held above the opening of the tube and stru

ck, while the far end of the tube is lengthened or shortened. Resonances (loudnesses) are heard at a series of tube lengths, L, which are carefully measured.
resonance tube

The speed of sound in air is determined by measuring the distance between two consecutive resonances. A data set for a tuning fork with a frequency of 320 Hz are L1 = 25.8 cm; L2 = 78.4 cm; and L3 = 131.1 cm. Based on these data, what is the speed of sound in air?

Hint: the opening at the top is an antinode while the water surface is a node. Therefore, what fraction of a full wave is L1? What fractions are L2 and L3?
Physics
1 answer:
LenaWriter [7]3 years ago
7 0

Answer:

330.24 Hz

Explanation:

Given:

Frequency, f = 320 Hz

L1 = 25.8 cm

L2 = 78.4 cm

L3 = 131.1 cm

Let the wavelength be λ

Then, L1 which is the length of the column of air is  λ/4.

λ/4 = 25.8 cm

λ = 25.8 × 4 = 103.2 cm = 1.032 m

Then, speed of sound in air is:

v = λ f

⇒ v = 1.032 × 320 Hz

⇒ v = 330.24 m/s

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Nastasia [14]

Explanation:

Given parameters:

Initial velocity = 72km/hr

Final velocity  = 0km/hr

Time taken  = 25s

Unknown:

Acceleration = ?

Solution:

To solve this problem, convert km/hr to m/s;

           1000m = 1km

           3600s = 1hr

  72km/hr;

          1km/hr  = 0.278m/s

         72km/hr = 0.278 x 72  = 20.02m/s

Acceleration is the change in velocity divided by the time taken;

       Acceleration = \frac{final velocity - initial velocity }{time}  

       Acceleration  = \frac{0 - 20.02}{25}   = -0.8m/s

The car  is actually decelerating at a rate of 0.8m/s

5 0
3 years ago
The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca
d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}

the elastic potencial energy is given by:

U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J

The work is defined as:

W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s

3 0
3 years ago
A sphere of mass of 1.55 kg is accelerated upwards by a string to which the sphere is attached. Its speed increases from 2.81 m/
Mama L [17]

Answer:

The tension in the string is 16.24 N

Explanation:

Given;

mass of the sphere, m = 1.55 kg

initial velocity of the sphere, u = 2.81 m/s

final velocity of the sphere, v = 4.60 m/s

duration of change in the velocity, Δt = 2.64 s

The tension of the string is calculated as follows;

T =  ma  + mg\\\\T = m(a + g)\\\\where;\\\\a \ is \ upward \ acceleration \ of \ the \ sphere\\\\g \ is \ acceleration \ due \ to \ gravity =9.8 \ m/s^2\\\\a = \frac{\Delta V}{\Delta t} = \frac{v- u}{ t} = \frac{4.6 - 2.81 }{2.64} = 0.678 \ m/s^2

T = 1.55(0.678 + 9.8)

T = 1.55(10.478)

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5 0
3 years ago
You hold a spherical salad bowl 50 cm in front of your face with the bottom of the bowl facing you. The salad bowl is made of po
nataly862011 [7]

Answer:

a) q = 39.29 cm ,  b)   h ’= - 3.929 cm  the image is inverted  and REAL

Explanation:

For this exercise we will use the equation of the constructor

          1 / f = 1 / p + 1 / q

where f is the focal length of the salad bowl, p and q are the distance to the object and the image

The metal salad bowl behaves like a mirror, so its focal length is

           f = R / 2

           f = 44/2

           f = 22 cm

a) Suppose that the distance to the object is p = 50 cm, let's find the distance to the image

           1 / q = 1 / f  - 1 / p

           1 / q = 1/22 - 1/50

           1 / q = 0.0254

            q = 39.29 cm

b) to calculate the size of the image we use the equation of magnification

           m = h’/ h = - q / p

            h ’= - q / p h

            h ’= - 39.29 / 50 5

            h ’= - 3.929 cm

the negative sign means that the image is inverted

as the rays of light pass through the image this is REAL

4 0
4 years ago
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Katarina [22]
<span>Strong nuclear forces hold the nucleus of an atom together. Weak nuclear forces are involved when certain types of atoms break down.</span>
3 0
3 years ago
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