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bonufazy [111]
3 years ago
13

Two tuning forks of frequency 480 hz and 484 hz are struck simultaneously. what is the beat frequency resulting from the two sou

nd waves?
Physics
1 answer:
umka21 [38]3 years ago
7 0
Ans: Beat frequency = f_b = 4Hz

Explanation: 
The beat frequency is equal to the absolute value of the difference in frequency of the two waves. In other words, the number of beats per second is equal to the difference in frequency. It is due to the destructive and constructive interference. <span>According to this interference, sound will be soft or loud.

Hence. the formula is:
</span>Beat frequency = f_b = |f_2 - f_1|
<span>
Since,
</span>f_1 = 480Hz
f_2 = 484Hz

Therefore,
Beat frequency = f_b = |484 - 480|

=> Beat frequency = f_b = 4Hz
-i
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3. Light travels from the Sun to Earth in 8.3 min. Given that the speed of light is 3.00108 m/s, what is the distance in meters
nasty-shy [4]

Answer:

13

Explanation:

13.0120481928 it is the distance

4 0
3 years ago
A 5.0 Ω resistor is hooked up in series with a 10.0 Ω resistor followed by a 20.0 Ω resistor. The circuit is powered by a 9.0 V
yan [13]
<h2>Answer:</h2>

(a) Attached to the response as Figure 1.

(b) 35.0Ω

(c) Across 5.0Ω = 1.3V

   Across 10.0Ω = 2.6Ω

   Across 20.0Ω = 5.2Ω

<h2>Explanation:</h2>

(a) The labelled circuit using the correct symbols (for the resistors and battery) has been attached to this response.

(b) Since the resistors are hooked up in series, their equivalent resistance R, is found by adding the individual resistances of the resistors (R₁, R₂ and R₃). i.e

R = R₁ + R₂ + R₃               -------------------(i)

Where;

R₁ = 5.0 Ω

R₂ = 10.0 Ω

R₃ = 20.0 Ω

<em>Substitute these values into equation (i) as follows;</em>

∴ R = 5.0 Ω + 10.0 Ω + 20.0 Ω

∴ R = 35.0 Ω

Therefore, the equivalent resistance is ∴ R = 35.0Ω

(c) When resistors are connected in series, the same current passes through them. To get the current through each resistor;

i. First, replace the resistors by their equivalent resistor as calculated above. The diagram has been attached to this response.

ii. As seen in the diagram, the current flowing through the equivalent resistor can be calculated using Ohm's law as follows;

V = I R              ------------------(ii)

Where;

V = Voltage supplied to the circuit = 9.0V

I = Current through the circuit

R = Resistance of the equivalent resistor = 35.0Ω

Substitute these values into equation (ii)

9.0 = I x 35.0

I = \frac{9.0}{35.0}

I = 0.26A

This is also the current flowing through each of the resistors separately.

iii. Calculate the voltage drop across

1.<em> 5.0 Ω resistor</em>

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 5.0Ω resistor

I = current through the 5.0Ω resistor = 0.26A

R = resistance of the 5.0Ω resistor = 5.0Ω

=> V = 0.26 x 5.0

=> V = 1.3V

2.<em> 10.0 Ω resistor</em>

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 10.0Ω resistor

I = current through the 10.0Ω resistor = 0.26A

R = resistance of the 10.0Ω resistor = 10.0Ω

=> V = 0.26 x 10.0

=> V = 2.6V

3.<em> 20.0 Ω resistor</em>

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 20.0Ω resistor

I = current through the 20.0Ω resistor = 0.26A

R = resistance of the 20.0Ω resistor = 10.0Ω

=> V = 0.26 x 20.0

=> V = 5.2V

7 0
3 years ago
A solid aluminum sphere of radius R has moment of inertia I about an axis through its center. What is the moment of inertia abou
frutty [35]

Answer:

I1 = 2/5 M1 R^2   for a sphere about its center

I2 = 2/5 M2 (2 R)^2 = 2/5 M2 R^ * 4 = 8/5 M2 R^2

Remember that M2 is greater than M1 by a factor 0f 2^3 = 8

Then I2 exceeds I1 by a factor of 32

6 0
3 years ago
When you put a pot of water on the stove, the stove transfers thermal energy to the water. As the water gains large
STALIN [3.7K]

Answer:

It releases some of the energy into the atmosphere as hot steam.

Explanation:

8 0
3 years ago
How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n
grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

dSin\theta = m\lambda

Where

m= Any integer which represent the number of repetition of spectrum

\lambda= Wavelength

d = Distance between the slits.

\theta= Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

d = \frac{m\lambda}{sin\theta }

d = \frac{2*536*10^{-9}}{sin(24)}

d = 2.63*10^{-6}m

Therefore the number of lines per millimeter would be given as

\frac{1}{d} = \frac{1}{2.63*10^{-6} }

\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})

\frac{1}{d} = 379.4 lines/mm

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

6 0
3 years ago
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