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olga_2 [115]
3 years ago
13

The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 400 g. The vibra

ting section of the string is 1.90 m long.What tension is needed to tune this string properly?Express your answer with the appropriate units.
Physics
1 answer:
Norma-Jean [14]3 years ago
4 0

Answer:

1456 N

Explanation:

Given that

Frequency of the piano, f = 27.5 Hz

Entire length of the string, l = 2 m

Mass of the piano, m = 400 g

Length of the vibrating section of the string, L = 1.9 m

Tension needed, T = ?

The formula for the tension is represented as

T = 4mL²f²/ l, where

T = tension

m = mass

L = length of vibrating part

F = frequency

l = length of the whole part

If we substitute and apply the values we have Fri. The question, we would have

T = (4 * 0.4 * 1.9² * 27.5²) / 2

T = 4368.1 / 2

T = 1456 N

Thus, we could conclude that the tension needed to tune the string properly is 1456 N

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A piston-cylinder device contains Helium gas initially at 150 kPa, 20 o C, and 0.5m 3 . The helium is now compressed in a polytr
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m=\frac{150\times 0.5}{2.076\times 293}

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From Energy balance

E_{in}-E_{out}=\Delta E_{system}

Neglecting kinetic and Potential Energy change

Q_{in}+W_{in}=change\ in\ Internal\ Energy

Change in Internal Energy \Delta U=u_2-u_1

\Delta U=mc_v(T_2-T_1)

\Delta U=0.123\times 3.115(140-20)

\Delta U=45.977 kJ

Q_{in}+57.48=45.977

Q_{in}=-11.50 kJ  

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