Which items in this picture provide insulation?

A sound wave leaves the loudspeaker. As it travels, it experiences a temporary increase in wavelength and then returns to its or

Brut [27]

A sound wave leaves the loudspeaker. As it travels, it experiences a temporary increase in wavelength and then returns to its original wavelength. The sound wave traveled through a helium balloon (helium is less dense than air could explain this change in wavelength

The pattern of disruption brought on by energy moving away from the sound source is known as a sound wave. Longitudinal waves are what makeup sound. This indicates that the direction of energy wave propagation and particle vibrational propagation are parallel. The atoms oscillate when they are put into vibration.

A high-pressure and a low-pressure zone are created in the medium as a result of this constant back and forth action. Compressions and rarefactions, respectively, are terms used to describe these high- and low-pressure zones. The sound waves go from one medium to another as a result of these regions being transmitted to the surrounding media.

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4 0

2 years ago

A +5.0 uC point charge is placed at the 0 cm mark of a meter stick and a -4.0 °C charge is

viva [34]

Answer:

-1.44*10^11v/m

Explanation:

6 0

3 years ago

A student sits on a rotating stool holding two 3.09-kg masses. When his arms are extended horizontally, the masses are 1.08 m fr

schepotkina [342]

Answer:

a

The New angular speed is $w_f = 2.034 rad/s$

b

The Kinetic energy before the masses are pulled in is $KE_i = 3.101 \ J$

c

The Kinetic energy after the masses are pulled in is $KE_f = 8.192 \ J$

Explanation:

From the we are told that masses are 1.08 m from the axis of rotation, this means that

The radius $r =1.08m$

The mass is $m = 3.09\ kg$

The angular speed $w = 0.770 \ rad/sec$

The moment of inertia of the system excluding the two mass $I = 3.25 \ kg \cdot m^2$

New radius $r_{new} = 0.34m$

Generally the conservation of angular momentum can be mathematical represented as

$w_f = [\frac{I_i}{I_f} ]w_i .....(1)$

Where $w_f$ is the final angular speed

$w_i$ is the initial angular speed

$I_i$ is the initial moment of inertia

$I_f$ is the final moment of inertia

Moment of inertia is mathematically represented as

$I = m r^2$

Where I is the moment of inertia

m is the mass

r is the radius

So the Initial moment of inertia is given as

$I_i = moment \ of \ inertia \ of\ the \ two \ mass \ + 3.25 \ kg \cdot m^2$

$I_i = 2m r^2 + 3.25$

The multiplication by is because we are considering two masses

$I_i = 2 [(3.09)(1.08)^2] +3.25 = 10.46 kg \cdot m^2$

So the final moment of inertia is given as

$I_f = 2[(3.09)(0.34)^2] +3.25 = 3.96 \ kg \cdot m^2$

Substituting these values into equation 1

$w_f = [\frac{10.46}{3.96} ] * 0.77 = 2.034 \ rad/sec$

Generally Kinetic energy is mathematically represented in term of moment of inertia as

$KE = \frac{1}{2} * I * w^2$

Now considering the kinetic energy before the masses are pulled in,

$KE_i = \frac{1}{2} * I_i * w^2_i$

The Moment of inertia would be $I_i = 10.46 \ Kg \cdot m^2$

The Angular speed would be $w_i = 0.77 \ rad/s$

Now substituting these value into the equation above

$KE_i = \frac{1}{2} * (10.46) * (0.770)^2 = 3.101 J$

Now considering the kinetic energy after the masses are pulled in,

$KE_f = \frac{1}{2} * I_f * w^2_f$

The Moment of inertia would be $I_f = 3.96 \ Kg \cdot m^2$

The Angular speed would be $w_f = 2.034 \ rad/s$

Now substituting these value into the equation above

$KE_f= \frac{1}{2} *(3.96)(2.034)^2$

$= 8.192J$

8 0

3 years ago

The parents of a young child carefully placed plastic plug protectors in all the electrical outlets in their house. Which statem

DanielleElmas [232]

I think the answer is c or b but i p ick c

4 0

3 years ago

Read 2 more answers

A 11.3-kg object oscillates at the end of a vertical spring that has a spring constant of 2.20 ✕ 104 N/m. The effect of air resi

3241004551 [841]

Answer:

(a) the frequency of the dampened oscillation is 7.02 Hz

(b) percentage decrease in amplitude of the oscillation in each cycle is 2%

Explanation:

Given;

mass of the object = 11.3 kg

the spring constant = 2.2 X 10⁴ N/m.

damping coefficient b = 3.00 N · s/m

Part (a) the frequency of the dampened oscillation

The oscillation frequency is calculated as follows;

$\omega _D = \sqrt{\omega_o^2 -(\frac{b}{2m})^2}\\\\\omega_o^2 = \frac{k}{m} =\frac{2.2X10^{4}}{11.3} = 1946.903rad/s\\\\thus, \omega _D = \sqrt{1946.903-(\frac{3}{2*11.3})^2} =44.12 rad/s$