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2 years ago

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[ b) The time of reverberation of an empty hall without and with 500 audiences is 1.5 sec and 1.4 sec respectively. Find the rev

erberation time with 800 audiences. Wait

1 answer:

Lilit [14]2 years ago

4 0

The reverberation time with 800 audiences is 0.875 seconds.

<h3>Reverberation time with 800 audience</h3>

R₁V₁ = R₂V₂

where;

R₁ is the reverberation time with 400 audience
R₂ is the reverberation time with 800 audience
V₁ is initial volume
V₂ is final volume

R₂ = R₁V₁/V₂

R₂ = (1.4 x 500) / 800

R₂ = 0.875 seconds

Thus, the reverberation time with 800 audiences is 0.875 seconds.

Learn more about reverberation time here: brainly.com/question/9278479

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What is amplitude, wavelength, and frequency?

krek1111 [17]

Wavelength- <span>distance between successive crests of a wave.

frequency- t</span><span>he rate at which something occurs or is repeated over time.

amplitude-</span><span> maximum extent of a vibration.</span>

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3 years ago

determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund

wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is $a_G= 6.78 m/s^2$

<u>Explanation</u>:

$N_A+N_B-Mg = 0$

$-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0$

$0.8N_B +0.8N_A = 975a_G$

$N_A+N_B = 9564.75$ -------------(1)

$-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0$

$-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0$

$-2.26N_A -0.06N_B= 0$ ----------------(2)

Solving the equation (1) and(2)

$N_A + N_B = 9564.75$

$-2.26N_A-0.06N_B=0$

$N_A = -260.85N$

$N_B = 9825.60N$

$\mu_s N_B + \mu_s N_A = 975a_G$

$0.8(9825.60)+0.8(-260.85) = 975a_G$ $a_G=\frac{7651.8}{975}$ $a_G_1=7.4848m/s^2$

Next lets assume that the front wheels contact with the ground N_A = 0

$F_B = Ma_G$

$N_B = M_g$

$N_B - M_g = 0$

$N_B(b-a) –F_Bh = 0$

$F_B = 975a_G$

$N_B-975(9.8) = 0$

$N_B=9564.75N$

$9564.75(2.20 -1.82) -F_B(0.55)=0$

$\frac{3634.605}{0.55}=F_B$

$F_B = 6608.3$

$F_B = Ma_G$

$6608.3 = 975a_G$

$a_G = 6.7778 m/s^2$

$a_G_2 = 6.78m/s^2$

Choosing the critical case

$a_G = min(a_G_1 ,a_G_2)$

$a_G = min(7.848, 6.78)$

$a_G= 6.78 m/s^2$

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4 years ago

If 478 watts of power are used in 14 seconds, how much work was done?

sergiy2304 [10]

" 478 watts " means 478 joules per second.

In 14 seconds, the total work or energy is (14 x 478) = <em>6,692 joules</em>

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3 years ago

In your own words explain how "Carrying Capacity" plays a vital role in today's world and our future as well

olganol [36]

Increased food production due to improved agricultural practices, control of many diseases by modern medicine and the use of energy to make historically uninhabitable areas of Earth inhabitable are examples of things which can extend carrying capacity.

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3 years ago

Read 2 more answers

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

LuckyWell [14K]

For a circular aperture, the first minima (n=1) as an angular separation from the peak of the central maxima given by

Sinθ = 1.22λ / d

Where,

d is the aperture or pupil diameter

d = 4.69 mm = 4.69 × 10^-3m

λ is the wavelength

λ = 545 nm = 545 × 10^-9 m

Then,

Sinθ = 1.22λ / d

Sinθ = 1.22 × 545 × 10^-9 / 4.69 × 10^-3

Sinθ = 1.418 × 10^-4 rad

Then, the head light sources have the same angular separation θ from the eye as the image have inside the eye.

For the headlight

Sinθ ≈ light separation / distantce for the eye

Light separation is give as x = 0.659 m

And let the distance of the eye be D

Then,

Sinθ = x / D

Make D subject of formula

D = x / Sinθ

D = 0.695 / 1.418 × 10^-4

D = 4902.316m

To km, 1km = 1000m

D ≈ 4.9 km

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3 years ago