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Novosadov [1.4K]
3 years ago
7

Each of three tuning forks, A, B, and C, has a slightly different frequency. When A and B are sounded together, they produce a b

eat frequency of 2 Hz. When A and C are sounded together, they produce a beat frequency of 5 Hz. What is the smallest possible beat frequency when B and C are sounded together
Physics
1 answer:
a_sh-v [17]3 years ago
7 0

Answer:

The smallest possible beat frequency when B and C are sounded together is 3 Hz

Explanation:

Given;

the beat frequency of A and B = 2 Hz

the beat frequency of A and C = 5 Hz

Beat frequency is equal to the difference in frequency of the two notes that interfere to produce the beats.

F_b_{(AB)} = 2 \ Hz\\\\F_A - F_B = 2 \ Hz -----equation(1)\\\\F_b_{(AC)} = 5 \ Hz\\\\F_A - F_C = 5 \ Hz-----equation(2)\\\\from \ equation(1), make \ F_A \ the \ subject \ of \ the \ formula\\\\F_A = 2H_z + F_B -----equation (3)\\\\Substitute \ in \ the \ value \ of \ F_A \ in \ equation(3) \ into \ equation(2)\\\\2H_z + F_B - F_C = 5H_z\\\\F_B-F_C = 5H_z - 2H_z\\\\F_B -F_C = 3Hz

Therefore, the smallest possible beat frequency when B and C are sounded together is 3 Hz

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A 3047.8 kg truck has lost its brakes coming down a mountain. Fortunately, there is a ramp of thick gravel inclined at 9.5 degre
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Answer:

The work done by the gravel to stop the truck is 520.44 kJ

Explanation:

<u>Step 1</u>: Data given

Mass of the truck = 3047.8 kg

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Velocity of  the truck = 20.68 m/s

distance = 26.6 meters

<u>Step 2:</u> Calculate initial kinetic energy

sin 9.5° = 0.165

h = ℓ*sin 9.5° = 26.6*0.165= 4.39 m

Ek = 1/2m*Vo² = 1/2*3047.8*20.68² = 651714.7 Joule = 651.7 kJ  = initial kinetic energy

<u>Step 3: </u>Calculate potential energy

Epot = U = m*g*h = 3047.8*9.81*4.39 = 131256.25 Joule = 131.26 kJ

<u>Step 4:</u>  What work is done by the truck on the gravel?  

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3 years ago
Suppose your surface body temperature averaged 90 degrees F. How much radiant energy in W/m^2 would be emitted from your body?
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493 \; \text{W}\cdot \text{m}^{-2}.

<h3>Explanation</h3>

The Stefan-Boltzmann Law gives the energy radiation <em>per unit area</em> of a black body:

\dfrac{P}{A} = \sigma \cdot T^{4}

where,

  • P the total power emitted,
  • A the surface area of the body,
  • \sigma the Stefan-Boltzmann Constant, and
  • T the temperature of the body in degrees Kelvins.

\sigma = 5.67 \times 10^{-8} \;\text{W}\cdot \text{m}^{-2} \cdot \text{K}^{-4}.

T = 90 \; \textdegree{}\text{F} = (\dfrac{5}{9} \cdot (90-32) + 273.15) \; \text{K} = 305.372 \; \text{K}.

\dfrac{P}{A} = \sigma \cdot T^{4} = 5.67 \times 10^{-8} \times 305.372^{4} = 493\; \text{W}\cdot \text{m}^{-2}.

Keep as many significant figures in T as possible. The error will be large when T is raised to the power of four. Also, the real value will be much smaller than 493\; \text{W}\cdot \text{m}^{-2} since the emittance of a human body is much smaller than assumed.

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