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Rashid [163]
3 years ago
11

Two pure tones are sounded together. The drawing shows the pressure variations of the two sound waves, measured with respect to

atmospheric pressure, where t1 = 0.0220 s and t2 = 0.0242 s. What is the beat frequency?
Physics
1 answer:
amm18123 years ago
7 0

Answer:

4.13Hz

Explanation:

f1 = 1/t1 = 1/0.022 = 45.45 Hz

f2 = 1/t2 = 1/0.0242= 41.32 Hz

No. of beats

= 45.45- 41.32

~ 4.13Hz

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How much energy is stored in the capacitor when it is aa fully charged
lakkis [162]

Answer:

0.5*10uF * 16*16 =0.0128

Explanation:

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2 years ago
What is the difference between a graph of linear motion and a graph of harmonic motion?
fredd [130]
Hope this helps :)

When describing linear motion, you need only one graph representing each of the three terms, while projectile motion requires a graph of the x and y axes. Graphs of simple harmonic motion are sine curves. Circular motion is different from other forms of motion because the speed of the object is constant.
5 0
3 years ago
n an experiment of a simple pendulum, measurements show that the pendulum has length m, mass kg, and period s. Take m/s2 . i. Us
barxatty [35]

Answer:

The answer is "(1.265 \pm 0.010) \ s \ and \ 0.709 \%"

Explanation:

In point i:

T_{theo}= 2\pi \sqrt{\frac{l}{g}}

        =2\pi\sqrt{\frac{0.397}{9.8}}\\\\= 1.265 \ s

If  error in the theoretical time period :

\frac{\Delta T_{theo}}{T_theo} = \frac{1}{2}  \frac{\Delta l }{l}\\\\\Delta T_{theo} = 1.265 \times \frac{1}{2} \times \frac{0.006}{0.397}

           = 0.010 \ s

 T_{theo} = (1.265 \pm 0.010) \ s

In point ii:

\% \ difference = \frac{|T_{exp} -T_{theo}|}{\frac{T_{exp}+T_{theo}}{2}} \times 100

<h3>                     = \frac{1.274 -1.265}{\frac{1.274+1.265}{2}} \times 100\\\\=0.709 \%</h3>
5 0
3 years ago
Develop an equation (with a proportionality constant) that describes the relationship between the gravitational force (fgrav), t
ivolga24 [154]

According to newton's law of gravitation, the gravitational force(F)  is directly proportional to the product mass of the moon(Mm) and the mass of the planet (Mp) and it is inversely proportional to the square of the separation between them.

Fg ∝ (Mp)(Mm) →(1)

Fg ∝ 1/d²→(2)

Combining equation (1) and (2),

Fg ∝ (Mp)(Mm)/d²

Fg =  G(Mp)(Mm)/d²

This is an equation that describes the relation between mass of moon (Mm) and mass of planet (Mp) and separation(d) between them.

To support the claim in favuor of this equation we use this equation to obtain the value of acceleration due to gravity on earth.

Let m be the mass of an object on earth then Fg between earth (Mp) and mass of an object is obtained by:

Fg =  G(Mp)(m)/R², where R= Radius of earth

This force is equal to the weight of an object i.e.,

g= G(Mp)/R²

Putting the values of G, Mp and R , we get, g=9.81 m/s²

which is the value we obtained on earth for acceleration due to gravity.

To know more about gravitational constant, visit:

brainly.com/question/13959861

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5 0
1 year ago
How to determine whether the object is magnetic.
o-na [289]

Answer:

By holding another magnet close to it. If the object is attracted to the magnet, then it too is magnetic.

7 0
3 years ago
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