All categories

3 years ago

2. What is the total effect of sound produced in an enclosed space called?

Physics

1 answer:

Igoryamba3 years ago

6 0

I believe it is Acoustics, the real name for it would be like Reverberation. But go with Acoustics. :)

You might be interested in

A student throws a set of keys vertically upward to his fraternity brother, who is in a window 3.60 m above. The brother's outst

Contact [7]

Answer:

$v_{i}=10.10 m/s$

Explanation:

The equation of the position is:

$y=y_{i}+v_{i}t-0.5gt^{2}$

Where:

v(i) is the initial velocity

The initial position y(i) will be zero and the final position y = 3.60 m.

So, we just need to solve this equation for v(i).

$v_{i}=\frac{y+0.5gt^{2}}{t}$

$v_{i}=\frac{3.6+0.5*9.81*1.6^{2}}{1.6}$

$v_{i}=10.10 m/s$

Therefore, the initial velocity is 10.10 m/s upwards.

I hope it helps you!

5 0

3 years ago

The question is in the attachment.

sammy [17]

Answer:

Correct answer: Yes, with acceleration a = 10 m/s²

Explanation:

By definition acceleration is equal.

a = ΔV / Δt = (V₂ - V₁) / (t₂ - t₁)

a = (10 - 0)/ (1 - 0) = (20 - 10)/ (2 - 1) = ........ (50 - 40)/ (5 -4) = 10 m/s²

a = 10 m/s²

God is with you!!!

5 0

4 years ago

if 2 houses are made of hardwood bit ones walls are twice as thick as the other, which house would be better at reducing the rat

Anestetic [448]

the one with thinner walls

3 0

3 years ago

If a woman needs an amplification of 1.0 ✕ 105 times the threshold intensity to enable her to hear at all frequencies, what is h

Damm [24]

Answer:

50 dB.

Explanation:

given,

Amplification needed by the woman,I= 1 x 10⁵ I₀

where as I₀ is the threshold frequency

overall hearing loss of the woman= ?

Using the equation of sound intensity level

$\beta(dB) = 10 log(\dfrac{I}{I_0})$

$\beta(dB) = 10 log(\dfrac{1\times 10^5\ I_0}{I_0})$

$\beta(dB) = 10 log(1\times 10^5)$

$\beta(dB) = 50\ dB$

The overall hearing loss of woman is equal to 50 dB.

7 0

4 years ago

A spring with a spring constant of 165 N/m is attached to a 2.0 kg mass and set into motion.

marin [14]

Explanation:

We have,

Spring constant of the spring, k = 165 N/m

Mass, m = 2 kg

It is required to find the period of the mass-spring system. For the spring mass system, the period is given by :

$T=2\pi \sqrt{\dfrac{m}{k} }\\\\T=2\pi \sqrt{\dfrac{2}{165} }\\\\T=0.69\ s$

The frequency of vibration is reciprocal of its time period. So,

$f=\dfrac{1}{T}\\\\f=\dfrac{1}{0.69}\\\\f=1.44\ Hz$

So, the period of the mass-spring system is 0.69 s and frequency is 1.44 Hz.

3 0

4 years ago