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ira [324]
2 years ago
7

The percent by which the fundamental frequency changed if the tension is increased by 30 percent is ? a)-20.04% b)-40.12% c)-30%

d)-14.02%
Physics
1 answer:
nika2105 [10]2 years ago
3 0

Answer:

Percentage increase in the fundamental frequency is

d)-14.02%

Explanation:

As we know that fundamental frequency of the wave in string is given as

f_o = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

now it is given that tension is increased by 30%

so here we will have

T' = T(1 + 0.30)

T' = 1.30T

now new value of fundamental frequency is given as

f_o' = \frac{1}{2L}\sqrt{\frac{1.30T}{\mu}}

now we have

f_o' = \sqrt{1.3}f_o

so here percentage change in the fundamental frequency is given as

change = \frac{f_o' - f_o}{f_o} \times 100

% change = 14.02%

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Read 2 more answers
A 1.0-μm-diameter oil droplet (density 900 kg/m3) is negatively charged with the addition of 39 extra electrons. It is released
GREYUIT [131]

Answer:

6.75\mu C/m^2

Explanation:

We are given that

Diameter,d=1\mu m=1\time 10^{-6} m

1\mu m=10^{-6} m

Radius,r=\frac{d}{2}=\frac{1}{2}\times 10^{-6}=0.5 \times 10^{-6} m

Density,\rho=900kg/m^3

Total number of electrons,n=39

Charge on electron =1.6\times 10^{-19} C

Total charge=q=ne=39\times 1.6\times 10^{-19}=62.4\times 10^{-19} C

Distance,s=2mm=2\times 10^{-3} m

Mass =density\times volume=900\times \frac{4}{3}\pi r^3=900\times \frac{4}{3}\pi(0.5\times 10^{-6})^3=4.7\times 10^{-16} kg

Initial velocity,u=0

Final speed,v=4.5 m/s

v^2-u^2=2as

(4.5)^2-0=2a(2\times 10^{-3})

20.25=4a\times 10^{-3}

a=\frac{20.25}{4\times 10^{-3}}=5062.5m/s^2

Force,F=ma

qE=ma

q(\frac{\sigma}{2\epsilon_0})=ma

\sigma=\frac{2\epsilon_0ma}{q}=\frac{2\times 8.85\times 10^{-12}\times 4.7\times 10^{-16}\times 5062.5}{62.4\times 10^{-19}}

\epsilon_0=8.85\times 10^{-12}

\sigma=6.75\times 10^{-6}C/m^2=6.75\mu C/m^2

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Answer:

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