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tino4ka555 [31]
2 years ago
13

Two closed organ pipe gives 4 beats when sounded together at 5°C. Calculate the number of beats in 35°C​

Physics
1 answer:
Gre4nikov [31]2 years ago
7 0

Answer:

The number of beats is 10.58 in 35°C.

Explanation:

The beat frequency is given by : f₁-f₂

At 5°C, f₁-f₂ = 4

We need to find the number of beats in 35°C​.

The frequency in a standing wave is proportional to \sqrt T.

So,

\dfrac{f_1-f_2}{f_1'-f_2'}=\dfrac{\sqrt {T}}{\sqrt{T'}}\\\\f_1'-f_2'=(f_1-f_2)\times \dfrac{\sqrt{T'}}{\sqrt{T}}\\\\f_1'-f_2'=4\times \dfrac{\sqrt{35}}{\sqrt{5}}\\\\f_1'-f_2'=10.58

So, the number of beats is 10.58 in 35°C.

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A small 1.0 kg steel ball rolls west at 3.0 m/s collides with a large 3.0 kg ball at rest. After the collision, the small ball m
77julia77 [94]

Answer:

The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

Explanation:

Given that,

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Mass of steel ball = 1.0 kg

Velocity = 3.0 kg

After collision,

Velocity = 2.0 m/s

Using conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

3.0\times0+1.0\times(3.0)(-i)=1.0\times2(-j)+3.0\times v_{2}

-3i+2j=3.0\times v_{2}

v_{2}=-i+0.66j

The direction of the momentum

tan\theta=\dfrac{0.66}{-1}

\theta=tan^{-1}\dfrac{0.66}{-1}

\theta=-33.42^{\circ}

The direction of the momentum with respect to east

\theta=180-33.42=146.58^{\circ}

Hence, The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

7 0
3 years ago
hat are the wavelengths of peak intensity and the corresponding spectral regions for radiating objects at (a) normal human body
bagirrra123 [75]

Answer:

(a)

\lambda _{m}=9.332 \times 10^{-6}m

(b)

\lambda _{m}=1.632 \times 10^{-6}m

(c) \lambda _{m}=4.988 \times 10^{-7}m

 

Explanation:

According to the Wein's displacement law

\lambda _{m}\times T = b

Where, T be the absolute temperature and b is the Wein's displacement constant.

b = 2.898 x 10^-3 m-K

(a) T = 37°C = 37 + 273 = 310 K

\lambda _{m}=\frac{b}{T}

\lambda _{m}=\frac{2.893\times 10^{-3}}{310}

\lambda _{m}=9.332 \times 10^{-6}m

(b) T = 1500°C = 1500 + 273 = 1773 K

\lambda _{m}=\frac{b}{T}

\lambda _{m}=\frac{2.893\times 10^{-3}}{1773}

\lambda _{m}=1.632 \times 10^{-6}m

(c) T = 5800 K

\lambda _{m}=\frac{b}{T}

\lambda _{m}=\frac{2.893\times 10^{-3}}{5800}

\lambda _{m}=4.988 \times 10^{-7}m

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