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den301095 [7]
3 years ago
6

In certain ranges of a piano keyboard, more than one string is tuned to the same note to provide extra loudness. For example, th

e note at 110 Hz has two strings at this frequency. If one string slips from its normal tension of 600 N to 540 N, what beat frequency is heard when the hammer strikes the two strings simultaneously?
Physics
1 answer:
Mama L [17]3 years ago
3 0

Answer:

f = 5.65 Hz

Explanation:

The fundamental frequency of a string is given by following formula:

f = v/2L

where,

f = fundamental frequency

v = speed of wave = √(TL/m)

L = Length of String

m = Mass of String

T = Tension in String

Therefore,

f = √(TL/m)/2L

2f = √(T/Lm)

For initial condition:

T₁ = 600 N

f₁ = 110 Hz

2(110 Hz) = √(600 N/Lm)

√(600 N)/220 Hz =  √Lm

Lm = 0.01239 N/s²

Now, for changed tension:

2f₂ = √(540 N/0.01239 N/s²)

f₂ = 208.7 Hz/2

f₂ = 104.35 Hz

So, the beat frequency will be:

f = f₁ - f₂

f = 110 Hz - 104.35 Hz

<u>f = 5.65 Hz</u>

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A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the
Maurinko [17]

Answer:

(a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity R_{0}=10\ mCi

Time t_{1}=4\ hours

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

R=R_{0}e^{-\lambda t}

\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})

Put the value into the formula

\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})

\lambda=0.0000154\ s^{-1}

\lambda=1.55\times10^{-5}\ s^{-1}

We need to calculate the half life

Using formula of half life

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}

Put the value into the formula

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}

T_{\dfrac{1}{2}}=44.719\times10^{3}\ s

T_{\dfrac{1}{2}}=11.3\ hr

(b). We need to calculate the value of N₀

Using formula of N_{0}

N_{0}=\dfrac{3.70\times10^{6}}{\lambda}

Put the value into the formula

N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}

N_{0}=2.38\times10^{11}\ nuclei

(c). We need to calculate the sample's activity

Using formula of activity

R=R_{0}e^{-\lambda\times t}

Put the value intyo the formula

R=10e^{-(1.55\times10^{-5}\times30\times3600)}

R=1.87\ mCi

Hence, (a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

4 0
3 years ago
Một chất điểm khối lượng m=200g chuyển
MrMuchimi
Can you please translate to English?
8 0
2 years ago
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