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irga5000 [103]
3 years ago
10

ransverse waves are sent along a 5.00-m-long string with a speed of 30.00 m/s. The string is under a tension of 10.00 N. What is

the mass of the string
Physics
1 answer:
yanalaym [24]3 years ago
7 0

Answer:

0.055 kg

Explanation:

According to the given situation the solution of the mass of the string is shown below:-

Speed of the wave is

v = \sqrt{\frac{F_T\times Length\ of\ string}{Mass\ of\ string}}

30.0 m/s = \sqrt{\frac{10 kg m/s^2\times 5.00 m}{Mass\ of\ string}

Mass of string is

= \sqrt{\frac{10 kg m^2/s^2\times 5.00 m}{900 m^2 s^2}

After solving the above equation we will get the result that is

= 0.055 kg

Therefore for calculating the mass of the string we simply applied the above formula.

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A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is
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Answer:

2856.96 J

0

0

\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f

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Kinetic energy

K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J

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P=mgy\\\Rightarrow P=mg0=0

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f

Half of maximum height

\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}

h_i=\frac{v_i^2}{2g}

v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s

The velocity of the athlete at half the maximum height is 6.78822 m/s

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