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3 years ago

Do two sound waves that seen equally loud always have the same amplitude? explain

1 answer:

3 0

No, because of that "seem" part of the question.

Your hearing system is not a linear sound detector and meter.
Sounds with equal amplitudes at different frequencies "seem"
to have different loudness, and the differences are different
for different people.

You've probably heard the saying "Things are not always as they seem."
The fact is, they NEVER are.

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A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total

Anna007 [38]

When it comes to this equation, I always stick to good ol' momma bears solutions! Tell that boy to put the rock down, rocks are dangerous!! From- MommaBoi101

6 0

3 years ago

An ideal gas is compressed at constant pressure p=1500 Pa from a volume Vi=0.4 m3 to Vf=0.25 m3. At the same time, heat is trans

Anon25 [30]

To solve this problem we will apply the first law of thermodynamics and we will make a balance between the heat transferred, its internal energy and the total work. Recall that for gases the definition of work can be expressed in terms of its pressure and volume. Let's start

$dQ = dU +dW$

Here,

dU = Internal Energy

dW = Work

But internal energy is unchanged, then

$dQ = dW$

$dQ = PdV$

Where

$dV$ = Change in Volume

P = Pressure

Finally, the expression of the heat transferred can be expressed in terms of pressure and volume, so it would end up becoming

$dQ = p(v_i-v_f)$

Replacing,

$dQ = (1500)(0.4-0.25)$

$dQ = 225J$

Therefore the correct answer is B.

3 0

4 years ago

Caroline, a piano tuner, suspects that a piano's G4 key is out of tune. Normally, she would play the key along with her G4 tunin

alexira [117]

Answer:

e. 425.9 Hz

Explanation:

The computation of the frequency is being played by the out-of-tune key is shown below;

Given that

Δf1 = x - 349.2 = 76.7.........(1)

Δf2 = 440 - x = 14.1......(2)

Now solve (1) and (2)

(440 - x) - x + 349.2 = 14.1 - 76.7

789.2 + (-2x) = -62.6

x = 425.9 Hz

Hence, the frequency is being played by the out-of-tune key is 425.9 Hz

Therefore the option e is correct

6 0

3 years ago

Which property of metals allows corrosion to take place? reactivity malleability ductility conductivity Description

horrorfan [7]

Your best bet would be reactivity

7 0

3 years ago

Read 2 more answers

In a "worst-case" design scenario, a 2000-{\rm kg} elevator with broken cables is falling at 4.00{\rm m/s} when it first contact

OLga [1]

Answer:

v=3.73m/s, $a=3.35m/s^{2}$

Explanation:

In order to solve this problem, we must first draw a diagram that will represent the situation (See attached picture).

So there are two ways in which we can solve this proble. One by using integrls and the other by using energy analysis. I'll do it by analyzing the energy of the system.

So first, we ned to know what the constant of the spring is, so we can find it by analyzing the energy on points A and C. So we get the following:

$K_{A}+U_{Ag}=K_{C}+U_{Cs}+U_{Cg}+W_{Cf}$

where K represents kinetic energy, U represents potential energy and W represents work.

We know that the kinetic energy in C will be zero because its speed is supposed to be zero. We also know the potential energy due to the gravity in C is also zero because we are at a height of 0m. So we can simplify the equation to get:

$K_{A}+U_{Ag}=U_{Cs}+W_{Cf}$

so now we can use the respective formulas for kinetic energy and potential energy, so we get:

$\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}ky_{C}+fy_{C}$

so we can now solve this for k, which is the constant of the spring.

$k=\frac{mV_{A}^{2}+mgh_{A}-fy_C}{y_{C}^{2}}$

and now we can substitute the respective data:

$k=\frac{(2000kg)(4m/s)^{2}+(2000kg)(9.8m/s^{2})(2m)-(17000N)(2m)}{(2m)^{2}}$

Which yields:

k= 9300N/m

Once we got the constant of the spring, we can now find the speed of the elevator at point B, which is one meter below the contact point, so we do the same analysis of energies, like this:

$K_{A}+U_{Ag}=K_{B}+U_{Bs}+U_{Bg}+W_{Bf}$

In this case ther are no zero values to eliminate so we work with all the types of energies involved in the equation, so we get:

$\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}mV_{B}^{2}+\frac{1}{2}ky_{B}^{2}+mgh_{B}+fy_{B}$

and we can solve for the Velocity in B, so we get:

$V_{B}=\sqrt{\frac{mV_{A}^{2}-ky_{B}^{2}+2mgh_{B}-2fy_{B}}{m}}$

we can now input all the data directly, so we get:

$V_{B}=\sqrt{\frac{(2000kg)(4m/s)^{2}-(9300N/m)(1m)^{2}+2(2000kg)(9.8m/s^{2})(1m)-2(17000N)(1m)}{(2000kg)}}$

which yields:

$V_{B}=3.73m/s$

which is the first answer.

Now, for the second answer we need to find the acceleration of the elevator when it reaches point B, for which we need to build a free body diagram (also included in the attached picture) and to a summation of forces:

$\sum F=ma$