The pitch of a note corresponds to which property of sound wave

The magnetic lines of force always travel from north to south true or false

kumpel [21]

Answer:

True

Explanation:

Magnetic field lines outside of a permanent magnet always run from the north magnetic pole to the south magnetic pole. Therefore, the magnetic field lines of the earth run from the southern geographic hemisphere towards the northern geographic hemisphere.

7 0

1 year 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

Scorpion4ik [409]

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:

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

4 0

2 years ago

Read 2 more answers

Select all that apply.

Jobisdone [24]

$\text{Acceleration}=\frac{d^2}{dt^2}\text{displacement}$
So the dimensions of acceleration is $LT^{-2}$
Any answer that comes under that definition is correct.

6 0

3 years ago

Read 2 more answers

In just 0.30 s , you compress a spring (spring constant 5000 n/m ), which is initially at its equilibrium length, by 4.0 cm. par

Ierofanga [76]

The average power output is the ratio between the work done to compress the spring, W, and the time taken, t:
$P= \frac{W}{t}$ (1)

The work done is equal to the elastic energy stored by the compressed spring:
$W=U= \frac{1}{2}kx^2$
where $k=5000 N/m$ is the spring constant and $x=4.0 cm=0.04 m$ is the compression of the spring. If we substitute the numbers, we find:
$W= \frac{1}{2}(5000 N/m)(0.04 m)^2=4 J$

And now we can use eq.(1) to calculate the average power output:
$P= \frac{W}{t}= \frac{4 J}{0.30 s} =13.3 W$

7 0

3 years ago

A parallel-plate capacitor is formed from two 9.1 cm-diameter electrodes spaced 1.3 mm apart. The electric field strength inside

vitfil [10]

Answer:

2.87nC

Explanation:

See attached file

6 0

2 years ago