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Neko [114]
3 years ago
6

Sound exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. Such a loudspeaker is mounted out

side on a pole. In winter, when the temperature is 273 K, the diffraction angle θ has a value of 19.5°. What is the diffraction angle for the same sound on a summer day when the temperature is 313 K?
Physics
1 answer:
blondinia [14]3 years ago
7 0

Answer:

\theta = 20.98 degree

Explanation:

As we know that the speed of the sound is given as

v = 332 + 0.6 t

now at t = 273 k = 0 degree

v = 332 m/s

so we have

a sin\theta = N\lambda

a sin\theta = N(\frac{v_1}{f})

now when temperature is changed to 313 K we have

t = 313 - 273 = 40 degree

now we have

v = 332 + (0.6)(40)

v_2 = 356 m/s

a sin\theta' = N(\frac{v_2}{f})

now from two equations we have

\frac{sin19.5}{sin\theta} = \frac{332}{356}

so we have

sin\theta = 0.358

\theta = 20.98 degree

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4*9 = 36 m/sec at 9 seconds. The average speed is (0 + 36)/2 = 18 .

Explanation:

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3 years ago
An electric drill transfers some energy wastefully. Aside from heat, what else carries the energy that it transfers wastefully?
earnstyle [38]

Sound and thermal/ heat energy.

Explanation:

Electric drill use electrical energy to convert it into kinetic, but there are some energy waste during the conversion in the form of sound and thermal caused by electric motor.

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3 years ago
The gravitational force between two objects that are 2.1 times 10^-1 m apart is 3.2 times 10^-6 N. If the mass of one object is
Shtirlitz [24]

Answer:

Mass of the other object is 38.45 kg.

Explanation:

Given:

The gravitational force between two objects is, F=3.2\times 10^{-6} N

Mass of one object is, m_{1}=55\ kg

Distance between the objects is, r=2.1\times 10^{-1}\ m

Gravitational constant is, G =6.674\times 10^{-11}\ m^3 kg^{-1}s^{-2}

Let the mass of the other object be m_{2}\ kg

Gravitational force is given as:

F=\frac{Gm_1m_2}{r^2}

Plug in the given values and solve for m_2. This gives,

3.2\times 10^{-6}=\frac{6.674\times 10^{-11}\times 55\times m_2}{(2.1\times 10^{-1})^2}\\3.2\times 10^{-6}=\frac{6.674\times 10^{-11}\times 55\times m_2}{0.0441}\\3.2\times 10^{-6}=8.3236\times 10^{-8}m_2\\m_2=\frac{3.2\times 10^{-6}}{8.3236\times 10^{-8}}= 38.45\ kg

Therefore, the mass of the other object is 38.45 kg.

3 0
4 years ago
Suppose 1 kg of Hydrogen is converted into Helium. a) What is the mass of the He produced? b) How much energy is released in thi
morpeh [17]

Answer:

a) m = 993 g

b) E = 6.50 × 10¹⁴ J

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we know atomic mass of helium = 4.002602

difference in the atomic mass of helium = 4.03176-4.002602 = 0.029158

fraction of mass lost = \dfrac{0.029158}{4.03176}= 0.00723

loss of mass for 1000 g = 1000 × 0.00723 = 7.23

a) mass of helium produced = 1000-7.23 = 993 g (approx.)

b) energy released in the process

E = m c²

E = 0.00723 × (3× 10⁸)²

E = 6.50 × 10¹⁴ J

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