Ask question

Ask question

All categories

2 years ago

11

You are trying to decide between two new stereo amplifiers. One is rated at 130 W per channel and the other is rated at 200 W pe

r channel. Part A In terms of dB, how much louder will the more powerful amplifier be when both are producing sound at their maximum levels?

1 answer:

NARA [144]2 years ago

7 0

Answer:

The sound level will be 1.870 dB louder.

Explanation:

Given that,

Power = 130 W

Power = 200 W

We need to calculate the sound level

Using formula of sound level

$I_{dB}=10\log(\dfrac{I}{I_{0}})$

For one amplifier,

$I_{1}=10\log(\dfrac{130}{I_{0}})$ ...(I)

For other amplifier,

$I_{2}=10\log(\dfrac{200}{I_{0}})$ ...(II)

For difference in dB levels

$I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}})-10\log(\dfrac{130}{I_{0}})$

$I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}}\times\dfrac{I_{0}}{130})$

$I_{2}-I_{1}=10\log(\dfrac{200}{130})$

$I_{2}-I_{1}=1.870\ dB$

Hence, The sound level will be 1.870 dB louder.

You might be interested in

A bicycle is ridden along a horizontal road with a driving force of 400, n,400n. its speed is constant at 12, m, slash, s,12m/s.

mr_godi [17]

The magnitude of the sum of the frictional forces acting on the bike and its rider is 400N.

<h3>What is friction force?</h3>

The friction force is the opposing force which acts on the object which is in relative motion.

The driving force is equal and opposite to the friction force acting between road and bicycle.

Friction force = 400N

The friction force between rider and bike is zero.

So the magnitude of sum of friction force = 400N +0 = 400N

Thus, the magnitude of the sum of the frictional forces acting on the bike and its rider.

Learn more about friction force.

brainly.com/question/1714663

#SPJ1

4 0

1 year ago

A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho

fiasKO [112]

Answer:

$v_f=8.17\frac{m}{s}$

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

$W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m$

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

$W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$

The box is initially at rest, so $v_i=0$ . Solving for $v_f$ :

$v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}$

5 0

2 years ago

viktelen [127]

Answer: D(t) = $8.e^{-0.4t}.cos(\frac{\pi }{6}.t )$

Explanation: A harmonic motion of a spring can be modeled by a sinusoidal function, which, in general, is of the form:

y = $a.sin(\omega.t)$ or y = $a.cos(\omega.t)$

where:

|a| is initil displacement

$\frac{2.\pi}{\omega}$ is period

For a Damped Harmonic Motion, i.e., when the spring doesn't bounce up and down forever, equations for displacement is:

$y=a.e^{-ct}.cos(\omega.t)$ or $y=a.e^{-ct}.sin(\omega.t)$

For this question in particular, initial displacement is maximum at 8cm, so it is used the cosine function:

$y=a.e^{-ct}.cos(\omega.t)$

period = $\frac{2.\pi}{\omega}$

12 = $\frac{2.\pi}{\omega}$

ω = $\frac{\pi}{6}$

Replacing values:

$D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)$

The equation of displacement, D(t), of a spring with damping factor is $D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)$ .

3 0

3 years ago

Which microscope has the highest potential magnification? comparison microscope scanning electron microscope compound microscope

TiliK225 [7]

The answer is the electron microscope. It is a kind of microscope that uses electrons to light a sample and make an enlarged image. These kinds of microscopes have much better determining power than light microscopes and can get much higher magnifications.

3 0

3 years ago

Read 2 more answers

1 m is equivalent to how many centimeters in the metric system

gladu [14]

100. A centimeter is a 100th of a meter.

5 0

2 years ago