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3 years ago

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It is found that a 5.70 m segment of a long string contains three complete waves and has a mass of 180 g. The string is vibratin

g sinusoidally with a frequency of 55.0 Hz and a peak-to-valley distance of 19.0 cm. (The "peak-to-valley" distance is the vertical distance from the farthest positive position to the farthest negative position). Calculate the wavelenght.

2 answers:

iragen [17]3 years ago

8 0

Answer:

wavelength = 3.8 m

Explanation:

As we know that linear mass density is defined as the ratio of mass and length

so here we have

$\mu = \frac{m}{L}$

$\mu = \frac{0.180}{5.70}$

now we have

$\mu = 0.0315 kg/m$

Now it is given that string contains three complete waves

length of one segment on string is half of the wavelength

so here we have

$3\frac{\lambda}{2} = 5.70 m$

$\lambda = 3.8 m$

So wavelength of the wave on string is 3.8 m

Hoochie [10]3 years ago

3 0

Answer:

1.9 m.

Explanation:

Three complete waves in the length of 5.7 m

The distance traveled by one complete wave is called wavelength.

Thus, the distance traveled by one wave = 5.7 / 3 = 1.9 m

Thus, the wavelength is 1.9 m.

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2 years ago

A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th

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1. 0.42 Hz

The frequency of a simple harmonic motion for a spring is given by:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k = 7 N/m is the spring constant

m = 1 kg is the mass attached to the spring

Substituting these numbers into the formula, we find

$f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz$

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

$T=\frac{1}{0.42 Hz}=2.38 s$

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: $x=0$ , so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

$E=K=\frac{1}{2}mv^2$

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

$E=U=\frac{1}{2}kA^2$

Since the total energy must be conserved, we have:

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m$

5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}kA^2$ is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m$

6. Maximum velocity: 1.17 m/s

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}mv_{max}^2$ is the mechanical energy of the system when x=0, which is when the system has maximum velocity, $v_{max}$

Equalizing the two expressions, we can solve to find $v_{max}$ , the maximum velocity:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m$

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2 years ago

To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm throug

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To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

$\tau = Fr$

Where,

F = Force

r = Radius

Replacing we have that,

$\tau = Fr$

$\tau = 21cm (\frac{1m}{100cm})* 550N$

$\tau = 11.55Nm$

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Finally, angular acceleration is a result of the expression of torque by inertia, therefore

$\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}$

$\alpha = \frac{11.55}{0.394}$

$\alpha = 29.3 rad/s^2$

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians $(\pi / 3)$ , therefore

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3 years ago

A capacitor is charged until its stored energy is 7.54 J. A second capacitor is then connected to it in parallel. If the charge

Ivan

Answer:

2 J

Explanation:

A charged capacitor of capacitance $C_1$ with energy of 7.54 J, is connected in parallel with another capacitor $C_2$ , so the charge is equally distributed between them.

(a) The energy stored in the capacitor before it being connected to the other capacitor is:

$U_O=q_0^2/2C_1=7.54 J\\$

The energy stored in the electric field is the sum of the energies of the two capacitors:

$U=U_1+U_2\\U=q_1^2/2C_1+q_2^2/2C_2$

since the charge equally distributed, $q_1$ = $q_2$ = $q_o/2$ . and since they are connected in parallel the potential difference on both of them is the same :

$V_1=V_2\\q_1/C_1=q_2/C_2\\q_0/C_1=q_0/C_2\\C_1=C_2=C_3\\$

hence,

$U=q_0^2/8C+q_0^2/8C\\U=q_0^2/4C\\U=2J$

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3 years ago