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Nitella [24]
3 years ago
14

A steel guitar string with a diameter of 0.300 mm and a length of 70.0 cm is stretched by 0.500 mm while being tuned. How much f

orce is needed to stretch the string by this amount? Young's modulus for steel is 2.0 × 1011 N/m2
Physics
1 answer:
Ganezh [65]3 years ago
8 0

Answer:

10.1 N

Explanation:

Your answer is 10.1 N, I don't actually know how to do it but I hope it helps.

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Two strings with linear densities of 5 g/m are stretched over pulleys, adjusted to have vibrating lengths of 0.50 m, and attache
HACTEHA [7]

Answer:

2.18 kg

Explanation:

The frequency of a wave in a stretched string f = n/2L√(T/μ) where n = harmonic number, L = length of string, T = tension = mg where m = mass of object on string and g = acceleration due to gravity = 9.8 m/s² and μ = linear density of string.

For string 1, its fundamental frequency f  is when n = 1. So,

f = 1/2L√(T/μ) =  1/2L√(mg/μ)

Now for string 1, L = 0.50 m, m = 20 kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f = 1/2L√(mg/μ)

f = 1/2 × 0.50 m√(20 kg × 9.8 m/s²/0.005 kg/m)

f = 1/1 m√(196 kgm/s²/0.005 kg/m)

f = 1/1 m√(39200 m²/s²)

f = 1/1 m × 197.99 m/s

f = 197.99 /s

f = 197.99 Hz

f ≅ 198 Hz

For string 2, at its third harmonic frequency f'  is when n = 3. So,

f' = 3/2L√(T/μ) =  3/2L√(mg/μ)

Now for string 2, L = 0.50 m, m = M kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f' = 3/2L√(Mg/μ)

f' = 3/2 × 0.50 m√(M × 9.8 m/s²/0.005 kg/m)

f' = 3/1 m√(M1960 m²/s²kg)

f' = 3/1 m√M√(1960 m²/s²kg)

f' = 3/1 m √M × 44.27 m/s√kg

f' = 132.81√M/s√kg

f' = 132.81√M Hz/√kg

Since the frequency of the beat heard is 2 Hz,

f - f' = 2 Hz

So, 198 Hz - 132.81√M Hz/√kg = 2 Hz

132.81√M Hz/√kg = 198 Hz - 2 Hz

132.81√M Hz/√kg = 196 Hz

√M Hz/√kg = 196 Hz/138.81 Hz

√M/√kg = 1.476

squaring both sides,

[√M/√kg] = (1.476)²

M/kg = 2.178

M = 2.178 kg

M ≅ 2.18 kg

8 0
3 years ago
The control group in an experimental study will __________.
Anettt [7]
Give you something to compare your results with. It's always nice to be able to see what changes have been made to the original, even if it's not technically the original (I know that was worded weird, I just don't know how else to explain it.) Hope this helped!
7 0
3 years ago
Read 2 more answers
A parallel-plate capacitor has an area of 4.59 cm2, and the plates are separated by 1.28 mm with air between them. it stores a c
nalin [4]
 <span>a) 
Capacitance = k x ε° x area / separation 
ε° = 8.854 10^-12 F/ m 
k = 2.4max 
average k = 0.78 / 1.27 * 2.4 +(1.27- 0.78) / 1.27 * 1 = 1.474 + 0.386 = 1.86 
(61.4 % separation k = 2.4 --- 38.6 % k = 1 air --- average k = 0.614 * 2.34 + 0.386 * 1 = 1.86 
area = 145 cm2 = 0.0145 m2 
separation = 1.27 cm 0.0127 m 

C = 1.86 * 8.854 10^-12 * 0.0145 / 0.0127 = 18.8 pF 
b) Q = C * V --- 18.8 * 83 = 1560.4 pC = 1.5604 nC 
c) E = V / d = 83 / 0.0127 = 6535.4 V/m </span>
7 0
3 years ago
What is meant by the motion of a force​
Vlad [161]

Answer:

Force is classified as a push or a pull

Explanation:

5 0
2 years ago
Is it true muscle mass affects someones flexibility
guapka [62]
It is very very true
8 0
4 years ago
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