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mrs_skeptik [129]

2 years ago

14

A stereo speaker is rated at P1000 = 52 W of output at 1000 Hz. At 20 Hz, the sound intensity level LaTeX: \betaβ decreases by 1

.3 dB. What is the power output P

1 answer:

baherus [9]2 years ago

3 0

Answer:

The value of the power is $P_c = 38.55 \ W$

Explanation:

From the question we are told that

The power rating $P_{1000} =P_b= 52 \ W$

The frequency is $f = 1000 \ Hz$

The frequency at which the sound intensity decreases $f_k = 20 \ Hz$

The decrease in intensity is by $\beta = 1.3 dB$

Generally the initial intensity of the speaker is mathematically represented as

$\beta_1 = 10 log_{10} [\frac{P_b}{P_a} ]$

Generally the intensity of the speaker after it has been decreased is

$\beta_2 = 10 log_{10} [\frac{P_c}{P_a} ]$

So

$\beta_1-\beta_2 = 10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]$

=> $\beta = 10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]= 1.3$

=> $\beta =10log_{10} [\frac{\frac{P_b}{P_a}}{\frac{P_c}{P_a}} ] = 1.3$

=> $\beta =10log_{10} [\frac{P_b}{P_c} ] = 1.3$

=> $10log_{10} [\frac{P_b}{P_c} ] = 1.3$

=> $log_{10} [\frac{P_b}{P_c} ] = 0.13$

taking atilog of both sides

$[\frac{P_b}{P_c} ] = 10^{0.13}$

=> $[\frac{52}{P_c} ] = 10^{0.13}$

=> $P_c = \frac{52}{1.34896}$

=> $P_c = 38.55 \ W$

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Sidana [21]

Answer:

The maximum height is 2881.2 m.

Explanation:

Given that,

Acceleration = 29.4 m/s²

Time = 7.00 s

We need to calculate the distance

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$s=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times29.4\times7^2$

$s=720.3\ m$

We need to calculate the velocity

Using formula of velocity

$v=a\times t$

Put the value into the formula

$v=29.4\times7$

$v=205.8\ m/s$

We need to calculate the height

Using formula of height

$H=\dfrac{v^2}{2g}$

Put the value into the formula

$H=\dfrac{(205.8)^2}{2\times9.8}$

$H=2160.9\ m$

We need to calculate the maximum height

Using formula for maximum height

$H'=H+s$

Put the value into the formula

$H'=2160.9+720.3$

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4 0

2 years ago

011 10.0 points

Ulleksa [173]

Answer:

2.47 m

Explanation:

Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.

The horizontal velocity of the ball is constant:

$v_x = v cos \theta = (25)(cos 35.9^{\circ})=20.3 m/s$

and the time taken to cover the horizontal distance d is

$t=\frac{d}{v_x}=\frac{52}{20.3}=2.56 s$

So this is the time the ball takes to reach the horizontal position of the crossbar.

The vertical position of the ball at time t is given by

$y=u_y t - \frac{1}{2}gt^2$

where

$u_y = v sin \theta =(25)(sin 35.9^{\circ})=14.7 m/s$ is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

And substituting t = 2.56 s, we find the vertical position of the ball when it is above the crossbar:

$y=(14.7)(2.56) - \frac{1}{2}(9.8)(2.56)^2=5.52 m$

The height of the crossbar is h = 3.05 m, so the ball passes

$h' = 5.52- 3.05 = 2.47 m$

above the crossbar.

8 0

2 years ago

Amanda spent 2/5 of her time after school doing homework and ¼ of her remaining time riding her bike. If she rode her bike for 4

Doss [256]

Answer:120 min

Explanation:

Given

Amanda spent $\frac{2}{5}$ of her time after school doing Home work

And $\frac{1}{4}$ of her remaining time riding her bike

It is given that she rode her bike for 45 minutes in a week

Let t be the time after school

therefore Amanda spend $\frac{2t}{5}$ in home work and $\frac{3t}{5}$ time is left

From remaining $\frac{3t}{5}$ time she spends $\frac{1}{4}$ time riding her bike

therefore $\frac{3t}{5}\times \frac{1}{4}=45$

thus $t=300 min$

therefore time spent on home work is $\frac{2}{5}\times 300=120 min$

6 0

2 years ago

Suppose that a rectangular toroid has 1,500 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i

V125BC [204]

Answer:

Current in the toroid will be $I=17.32\times 10^{-3}A$

Explanation:

We have given number of winding in rectangular toroid N = 1500

Self inductance of toroid L = 0.06 H

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We have to find the current in the toroid

Magnetic energy stored is equal to $E=\frac{1}{2}Li^2$

$9\times 10^{-6}=\frac{1}{2}\times 0.06\times I^2$

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So current in the toroid will be $I=17.32\times 10^{-3}A$

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3 years ago

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katrin2010 [14]

Answer:

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23.4843749996 m

Yes

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q = Charge of electron = $1.6\times 10^{-19}\ C$

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Dividing by g

$\dfrac{a}{g}=\dfrac{1.9161676647\times 10^{13}}{9.81}\\\Rightarrow a=1.9532799844\times 10^{12}g$

The acceleration is $1.9161676647\times 10^{13}\ m/s^2$ or $1.9532799844\times 10^{12}g$

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The distance is 23.4843749996 m

The gravitational field is very small compared to the electric field so the effects of gravity can be ignored.

5 0

2 years ago