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skad [1K]
3 years ago
7

Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our

last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (at 76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F ?
Physics
1 answer:
Gnom [1K]3 years ago
3 0

Answer:

 f = v / 4L

the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.

Explanation:

In wind instruments the wave speed must meet

          v = λ f

          λ = v / f

from v is the speed of sound that depends on the temperature

          v = v₀ \sqrt{1+ \frac{T [C]}{273} }

where I saw the speed of sound at 0ºC v₀ = 331 m/s  the temperature is in degrees centigrade, we can take the degrees Fahrenheit to centigrade with the relation

            (F -32) 5/9 = C

            76ºF = 24.4ºC

            45ºF = 7.2ºC

           

With this relationship we can see that the speed of sound is significantly reduced when leaving the house to the outside

at T₁ = 24ºC               v₁ = 342.9 m / s

at T₂ = 7ºC                 v₂ = 339.7 m / s

To satisfy this speed the wavelength of the sound must be reduced, so the resonant frequencies change

              λ / 4 = L

              λ= 4L

              v / f = 4L

              f = v / 4L

Therefore, the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.

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A positive charge of 8.0 × 10-4 C is in an electric field that exerts a force of 3.5 × 10-4 N on it. What is the strength of the
Gennadij [26K]

Answer:

E = 0.437 N/C

Explanation:

Given that,

Charge, q=8\times 10^{-4}\ C

Electric force, F=3.5\times 10^{-4}\ N

Let the strength of the electric field is E. We know that, the electric force is given by :

F = qE

Where

E is the electric field strength

E=\dfrac{F}{q}\\\\E=\dfrac{3.5\times 10^{-4}}{8\times 10^{-4}}\\E=0.437\ N/C

So, the strength of the electric field is equal to 0.437 N/C.

6 0
3 years ago
A balloon filled with helium gas has an average density of rhob = 0.22 kg/m3. The density of the air is about rhoa = 1.23 kg/m3.
MissTica

Answer:

a=g\left(\frac{\rho_a}{\rho_b}-1\right)

45.03681 m/s²

Explanation:

F_b = Buoyant force

W = Weight of the balloon

\rho_a = Density of air = 1.23 kg/m³

\rho_b = Density of balloon = 0.22 kg/m³

v_a = Volume of air

v_b = Volume of balloon

F_b=\rho_av_bg

W=\rho_bv_bg

g = Acceleration due to gravity = 9.81 m/s²

The net force acting on the balloon is

F=F_b-W\\\Rightarrow F=\rho_av_bg-\rho_bv_bg\\\Rightarrow \rho_bv_ba=\rho_av_bg-\rho_bv_bg\\\Rightarrow \rho_bv_ba=v_bg(\rho_a-\rho_b)\\\Rightarrow a=\frac{g}{\rho_b}(\rho_a-\rho_b)\\\Rightarrow a=g\left(\frac{\rho_a}{\rho_b}-1\right)

The equation is a=g\left(\frac{\rho_a}{\rho_b}-1\right)

a=g\left(\frac{\rho_a}{\rho_b}-1\right)\\\Rightarrow a=9.81\times \left(\frac{1.23}{0.22}-1\right)\\\Rightarrow a=45.03681\ m/s^2

The acceleration of the balloon is 45.03681 m/s²

8 0
3 years ago
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laiz [17]

Answer:

a toy car was on the floor in the play room until the child applied force by starting to race with it. it accelerated and proceeded to move at 5m/s east. it came to a stop when the bedpost applied force to it.

7 0
3 years ago
FOR EACH GROUP OF TERMS, WRITE A SENTENCE THAT SHOWSHOW THE TERMS ARE RELATED:
skad [1K]
1. A body is in motion when it changes its position at a certain time/duration.
2. Speed is related to how fast a body is moving.
3. Speed means that a body is moving and covering distance at a certain time.
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5 0
4 years ago
A 5 kg block moves with a constant speed of 10 ms to the right on a smooth surface where frictional forces are considered to be
nirvana33 [79]

Answer:

Work done, W = 19.6 J

Explanation:

It is given that,

Mass of the block, m = 5 kg

Speed of the block, v = 10 m/s

The coefficient of kinetic friction between the block and the rough section is 0.2

Distance covered by the block, d = 2 m

As the block passes through the rough part, some of the energy gets lost and this energy is equal to the work done by the kinetic energy.

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So, the change in the kinetic energy of the block as it passes through the rough section is 19.6 J. Hence, this is the required solution.

5 0
3 years ago
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